मंथिकाManthika
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From triangle ratios to circle functions

You already know sine and cosine as ratios of a right triangle’s sides. Take a triangle with an angle of 30 degrees: the side opposite it, divided by the hypotenuse, is $1/2$. That is $\sin(30°)$.

Now ask for $\sin(200°)$. The definition stops working. No right triangle has an angle of 200 degrees, because its three angles must add to 180. There is no triangle to take a ratio from.

So move the angle somewhere with more room. Draw a circle of radius 1 centred at the origin. Measure the angle from the positive x-axis, turning anticlockwise, and mark where the turn lands on the circle. That point has coordinates. Call them $\cos \theta$ and $\sin \theta$.

Check that nothing broke. For an angle between 0 and 90 degrees the point sits in the first quadrant. Drop a line from it to the x-axis and you have a right triangle with hypotenuse 1. So the coordinates ARE the old ratios. Same values, new definition.

What the new definition buys is room. The angle can now be 200 degrees, or 500, or negative. The point still lands somewhere, and it still has coordinates.

The ratio moved off the triangle and onto the circle. That is what turns a ratio into a function.

Nothing happens to the curve at 90 degrees — no break, no kink, no change of rule. The limit is in the old definition, not in the function, and that is why extending the definition costs nothing and buys the whole line.
The right triangle only exists in the first quadrant. The figure keeps it there so you can see that its two legs are also the point’s two coordinates — nothing is being redefined when the angle grows. Past 90 degrees there is no acute angle left to take a ratio of, yet the point still has an x and a y, and at 140 degrees the x is negative. A ratio of two lengths can never be negative; a coordinate can, and that is what moving onto the circle adds.

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Measuring the angle: degrees and radians

▲CONCEPT

Degrees are not the only way to measure an angle, and past this chapter they are not the usual way.

The whole conversion rests on one fact. A full circle is 360 degrees, and it is also $2 \pi$ radians. Halve both: $\pi$ radians is exactly 180 degrees.

From that, both directions follow. To convert DEGREES to RADIANS, multiply by $\pi/180$. To convert RADIANS to DEGREES, multiply by $180/\pi$.

If you ever forget which way round it goes, rebuild it instead of guessing. The factor you want is the one whose units cancel. Put 180 underneath when you start from degrees, and the degrees divide out.

*One fact, $\pi$ radians is 180 degrees, generates both conversions. Memorise the fact, not the two formulas.*

The two scales are on one ruler, not on two, and that is the whole content of the conversion. Put a finger on 135 degrees along the lower edge, follow the line straight up, and the upper edge reads three quarters of pi. The finger never moved, so the angle never changed. Only the far end has to be memorised: half a turn is 180 degrees below and pi above. Everything else on the ruler is that one agreement rescaled, and the two multipliers are only that same ratio written the two ways round.
Worked example

Convert 45 degrees to radians, and $5 \pi/6$ radians to degrees

  1. $45 \times \pi/180 = \pi/4$
    going from degrees, multiply by $\pi/180$
  2. 90 degrees $= \pi/2$
    check against something known — 90 degrees is a quarter turn, and 45 degrees is half of that, so $\pi/4$ is right
  3. $(5 \pi/6) \times (180/\pi)$
    going the other way, multiply by $180/\pi$
  4. $5 \times 30 = 150$ degrees
    the $\pi$ on top cancels the $\pi$ underneath, leaving $180/6 = 30$
  5. no $\pi$ remains
    radian answers usually carry a $\pi$, degree answers never do — that cancellation is what removes it; a degree answer with a $\pi$ still in it is a signal you multiplied the wrong way round
Each conversion is one angle wearing two names, not two quantities. The left sweep is an eighth of the way round and the right one is five sixths of the way to the far end. The sizes are settled by the picture before any arithmetic runs --- an answer that put 5 pi over 6 anywhere near the left panel’s sweep would be visibly wrong. All the algebra underneath does is rename the same sweep on the other scale.
▲CONCEPT

Radians are not just another unit. They are chosen so that one particular formula comes out clean.

For a circle of radius r, an arc that subtends an angle $\theta$ at the centre has length $l = r \theta$.

That is as simple as such a formula gets: no constant in front, no conversion factor. Multiply the radius by the angle and you have the arc length.

It only works in RADIANS. Put degrees into $l = r \theta$ and the answer is wrong, because degrees carry an arbitrary 360-per-circle in them. Radians are defined the other way round — defined precisely so that this formula needs no correction.

That is the honest reason radians exist. They are the measure that makes angle and arc length the same kind of thing.

*Check the units before using $l = r \theta$. In degrees it is simply false.*

Compare the first two bars. The arc cut off by one radian is exactly as long as the radius, and that is not a result to be proved — it is what one radian means. The third bar is 2.5 times the first because the angle is 2.5 radians. Measure the angle this way and the arc length is simply the radius multiplied by the angle, with no conversion factor left over. Measure it in degrees and you have to carry one.

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Signs, quadrant by quadrant

The unit-circle point has coordinates, and coordinates have signs. Everything about which functions are positive where follows from that, so it does not need separate memorising.

Cosine is the x-coordinate, and sine is the y-coordinate. Tangent is sine over cosine, so its sign is the two signs divided.

Quadrant by quadrant.

Everything not named is negative in that quadrant.

Work it from the coordinate signs. Then the rule is derived each time rather than recalled wrongly.

The four points sit the same distance from both axes; the only thing that changes is which side of the origin they fall on. Read each one as the pair (cos, sin) and the first two rows of the rule follow at once. The sign of the cosine is the sign of x, the sign of the sine is the sign of y. The tangent is the one that needs a second look. It is y divided by x, so it is positive wherever x and y carry the same sign — the first quadrant and the third.
Worked example

Find the quadrant where $\sin(\theta) < 0$ and $\cos(\theta) < 0$

  1. $\sin(\theta) < 0$: Quadrants III, IV
    sine is the y-coordinate, negative below the x-axis
  2. $\cos(\theta) < 0$: Quadrants II, III
    cosine is the x-coordinate, negative to the left of the y-axis
  3. Quadrant III
    the only quadrant in both lists — matches the picture directly, since Quadrant III is the bottom-left, where x and y are both negative; two sign conditions, two short lists, one overlap: the standard shape of a quadrant question
Each condition on its own rules out only half the circle, and that is the step the written answer skips. Sine negative leaves the whole bottom, cosine negative leaves the whole left. Only where the two shaded regions lie on top of each other are both true at once, and that is the third quadrant alone. The point at 225 degrees is there so the answer can be checked rather than taken: both of its coordinates land on the negative side of their own axis.

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The values worth knowing cold

Five angles come up so often that their exact values are worth knowing cold: 0, 30, 45, 60 and 90 degrees.

Sine takes these values, in that order: $0$, $1/2$, $1/\sqrt{2}$, $\sqrt{3}/2$, $1$.

Cosine takes the same five values in REVERSE order: $1$, $\sqrt{3}/2$, $1/\sqrt{2}$, $1/2$, $0$.

That reversal is not a coincidence to be memorised separately. It is $\sin(90 - \theta) = \cos(\theta)$ at work. Reading the sine list backwards is the same as replacing each angle by 90 minus itself, which turns every sine into a cosine.

Tangent is sine over cosine. Divide the entries: $0$, $1/\sqrt{3}$, $1$, $\sqrt{3}$, and at 90 degrees nothing, because cosine is 0 there and tangent is not defined.

Learn the sine row. The cosine row is it backwards, and the tangent row is one divided by the other.

The line y = x is a mirror, and reflecting a point in it swaps that point’s two coordinates. On the circle the same reflection carries 30 degrees to 60 degrees. This is why the sine of one is the cosine of the other, and why the cosine row of the table is the sine row read backwards. Forty-five degrees lies on the mirror itself, so it is its own reflection, and there the sine and the cosine are equal.

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Turning the other way — odd and even

▲CONCEPT

Turning by $-\theta$ instead of $\theta$ means turning the same amount the other way. On the circle, that reflects the point across the x-axis.

Reflecting across the x-axis leaves the x-coordinate alone and flips the sign of the y-coordinate.

Cosine is the x-coordinate, so it is unchanged: $\cos(-\theta) = \cos(\theta)$. A function behaving this way is called EVEN.

Sine is the y-coordinate, so it flips: $\sin(-\theta) = -\sin(\theta)$. A function behaving this way is called ODD.

Tangent follows from the two, being sine over cosine. The top flips and the bottom does not, so the ratio flips with it. That gives $\tan(-\theta) = -\tan(\theta)$. Tangent is odd.

One reflection, read off two coordinates. Cosine survives it. Sine does not.

Fold the cosine graph along the vertical axis and it lands on itself: −2π/3 and 2π/3 give the same height, −1/2. Turn the sine graph half a turn about the origin and it lands on itself: the two heights are equal and opposite, so the line joining them passes through the origin.
Both angles reach the horizontal axis at the same point, so their x-coordinates are one number and not two: the cosine cannot tell the two angles apart. The y-coordinates are equal and opposite, so the sine changes sign. Follow the two rays out to the line on the right and the tangent changes sign as well, because it is y divided by x and only the y moved.

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The Pythagorean identities

▲CONCEPT

The unit-circle point sits at distance 1 from the origin, because the radius is 1. Write that distance out with Pythagoras and the chapter’s central identity appears: $\sin^2(\theta) + \cos^2(\theta) = 1$, for every angle $\theta$.

Two more identities come from that one, and both are worth deriving rather than memorising.

Divide every term by $\cos^2(\theta)$. The first term becomes $\tan^2(\theta)$, the second becomes 1, and the right-hand side becomes $\sec^2(\theta)$. Rearranged, that is $1 + \tan^2(\theta) = \sec^2(\theta)$.

Now divide the original by $\sin^2(\theta)$ instead. The same move gives $1 + \cot^2(\theta) = \csc^2(\theta)$.

All three hold wherever the functions in them are defined. The second and third fail only where their division was illegal in the first place.

One identity and two divisions. If you can do the divisions, you never have to recall the other two.

Read the identity backwards and it becomes a tool: knowing either square hands you the other by subtraction. Taking the square root then returns two candidate signs, and it is the quadrant — not this identity — that picks between them.
The small triangle is Pythagoras on the unit circle, and its hypotenuse is the radius — which is why the two squares add to exactly 1. The other two are that same triangle enlarged. Dividing the identity through by the square of the cosine, and enlarging the triangle until its bottom leg measures 1, are the same act done in two notations. Divide by the square of the sine instead and it is the upright leg that becomes 1.
Worked example

Find $\cos(\theta)$ given $\sin(\theta) = 3/5$ in Quadrant II

  1. $\sin^2(\theta) + \cos^2(\theta) = 1$
    start from the Pythagorean identity
  2. $\cos^2(\theta) = 1 - 9/25 = 16/25$
    substitute $\sin^2(\theta) = 9/25$ and rearrange
  3. $\cos(\theta) = \pm 4/5$
    take the square root — both signs satisfy the identity, so the identity alone cannot finish the job
  4. $\cos(\theta) = -4/5$
    the quadrant decides — in Quadrant II cosine is negative; the identity gives the size, the quadrant gives the sign, and skipping that step leaves only a 50-50 chance of the right answer
One horizontal line meets the circle twice, and that second crossing is the plus-or-minus the square root produces --- it is on the page precisely because it is the answer this question rejects. Everything the algebra can know stops at step 3. What settles the sign is where the foot of the triangle lands: on the negative side of the x-axis, because the angle was given as a second-quadrant angle.

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Combining angles

▲CONCEPT

Knowing the ratios of A and of B does not immediately give the ratios of $A+B$. Four formulas do.

For cosine: $\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$, and $\cos(A-B) = \cos(A) \cos(B) + \sin(A) \sin(B)$.

For sine: $\sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B)$, and $\sin(A-B) = \sin(A) \cos(B) - \cos(A) \sin(B)$.

Two details catch people out. The cosine formulas carry the OPPOSITE sign to the one in the bracket — a plus inside gives a minus outside. The sine formulas carry the SAME sign. And every formula mixes the two functions: no correct one is built from sines alone or cosines alone.

The sign flips for cosine and holds for sine. Getting that backwards is the most common error here.

The right-hand circle is not a second picture, it is the left one turned. Nothing has been stretched. The chord that was there before is the same chord now --- and it is now stretched between a point at (1, 0) and a point at a single angle A minus B. Writing that one length two ways and squaring is the entire derivation. The products of cosines and of sines that the formula asks you to memorise are what falls out when the brackets are opened.
Worked example

Find the exact value of $\sin(75°)$

  1. $75 = 45 + 30$
    75 is not a standard angle, but it is a sum of two that are — split an awkward angle into standard ones
  2. $\sin(45°+30°) = \sin(45°) \cos(30°) + \cos(45°) \sin(30°)$
    apply the sine sum formula
  3. $\sin(45°) = 1/\sqrt{2}$, $\cos(45°) = 1/\sqrt{2}$
    the standard values at 45 degrees
  4. $\cos(30°) = \sqrt{3}/2$, $\sin(30°) = 1/2$
    the standard values at 30 degrees
  5. $(1/\sqrt{2})(\sqrt{3}/2) + (1/\sqrt{2})(1/2)$
    substitute the standard values into the sum formula
  6. $(\sqrt{3} + 1)/(2 \sqrt{2})$
    both terms share $1/\sqrt{2}$ — factor it out and add what is left; nothing was measured or approximated, an exact value came entirely out of two standard angles
The step worth learning here is the first one, and it is a step about noticing rather than calculating. 75 is not a standard angle, but it is 45 with 30 stacked on top, and both of those you know cold. The two arcs are drawn at different radii so the stacking is a shape and not a claim. Reading sine 75 as a height also gives you a check the algebra cannot: it must come out just under 1, and a little above sine 45.
▲CONCEPT

The double-angle formulas are not new results. They are the sum formulas with B set equal to A.

Put $B = A$ into the sine sum formula and the two terms become identical, so they add: $\sin(2A) = 2 \sin(A) \cos(A)$.

Do the same to the cosine sum formula: $\cos(2A) = \cos^2(A) - \sin^2(A)$.

That cosine result has two more forms, and they matter. Use $\sin^2(A) + \cos^2(A) = 1$ to replace $\sin^2(A)$ by $1 - \cos^2(A)$, and you get $\cos(2A) = 2 \cos^2(A) - 1$. Replace $\cos^2(A)$ instead and you get $\cos(2A) = 1 - 2 \sin^2(A)$.

All three are the same identity. Which one to reach for depends on what you already have. If you know cosine, use the version in cosines only. If you know sine, use the version in sines only.

Dividing $\sin(2A)$ by $\cos(2A)$, and then dividing every term by $\cos^2(A)$, gives tangent’s own version: $\tan(2A) = (2 \tan(A))/(1 - \tan^2(A))$.

One identity, three faces. Pick the face that matches what you were given.

Stacking the same angle twice is exactly what setting B equal to A means, and drawing it settles the question the formula is really answering. Doubling the angle does not double the height: twice sine A lands above the circle here, at a value no sine can take. That is why sine 2A carries a cosine A factor rather than a 2. Why the three forms of cosine 2A are one identity read three ways rather than three things to learn.
Worked example

Find $\sin(2 \theta)$ given $\sin(\theta) = 3/5$ in Quadrant I

  1. $\sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$
    the double-angle formula — sine is given, so cosine is the missing piece
  2. $\cos^2(\theta) = 1 - 9/25 = 16/25$, so $\cos(\theta) = 4/5$
    get cosine from the Pythagorean identity — Quadrant I has everything positive, so take the positive root
  3. $\sin(2 \theta) = 2 (3/5)(4/5) = 24/25$
    substitute both values into the formula
  4. $2 \sin(\theta) = 6/5$
    NOT the answer — a value above 1 is impossible for a sine, proof that doubling the angle is not doubling the sine

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Tangent and cotangent of a sum

▲CONCEPT

The sum and difference formulas give sine and cosine. Tangent’s own version is not a separate fact — it falls straight out of the two you already have.

Divide the sine sum formula by the cosine sum formula, then divide every term, top and bottom, by $\cos(A) \cos(B)$: $\tan(A+B) = (\tan(A) + \tan(B))/(1 - \tan(A) \tan(B))$. The same division on the difference formulas gives $\tan(A-B) = (\tan(A) - \tan(B))/(1 + \tan(A) \tan(B))$.

Cotangent is the reciprocal of tangent, so inverting these gives cotangent’s versions: $\cot(A+B) = (\cot(A) \cot(B) - 1)/(\cot(B) + \cot(A))$, and $\cot(A-B) = (\cot(A) \cot(B) + 1)/(\cot(B) - \cot(A))$.

Notice the pattern from the double-angle work repeats: a sum on top, a product-minus-one or one-minus-product underneath, and the sign in the denominator always opposite the sign in the numerator.

Four more formulas, zero new facts — every one of them is the sine and cosine sum/difference formulas, divided through.

Each tangent value is a height on the line x = 1. The same 15° turn cuts a step of 0.268 at the bottom but 0.423 between 30° and 45°, so tan 45° − tan 30° is not tan 15°. Dividing that step by 1 + tan 45° tan 30° = 1.577 gives the true value, 2 − √3.
Worked example

Find the exact value of $\tan(15°)$

  1. $15 = 45 - 30$
    15 is not a standard angle, but it is a difference of two that are
  2. $\tan(45°-30°) = (\tan(45°) - \tan(30°))/(1 + \tan(45°)\tan(30°))$
    apply the tangent difference formula
  3. $\tan(45°) = 1$, $\tan(30°) = 1/\sqrt{3}$
    the standard values at 45 and 30 degrees
  4. $(1 - 1/\sqrt{3})/(1 + 1/\sqrt{3}) = (\sqrt{3} - 1)/(\sqrt{3} + 1)$
    substitute the standard values, then multiply numerator and denominator by $\sqrt{3}$
  5. $2 - \sqrt{3}$
    multiply numerator and denominator by $(\sqrt{3} - 1)$ and simplify — an exact value out of two standard angles, same as the sine and cosine worked examples
A second route to the same answer: extend the 30° triangle’s base by its own hypotenuse, 2, and the angle at the far end is 15°, so tan 15° = 1/(2 + √3) = 2 − √3.

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Three times the angle

▲CONCEPT

Double-angle formulas came from $B = A$ in the sum formula. Triple-angle formulas come from writing $3x$ as $x + 2x$ and applying the sum formula again.

$\sin(3x) = \sin(x + 2x) = \sin(x) \cos(2x) + \cos(x) \sin(2x)$. Substitute the double-angle formulas for $\cos(2x)$ and $\sin(2x)$, then use $\sin^2(x) + \cos^2(x) = 1$ to remove every leftover $\cos^2(x)$, and the result simplifies to one function only: $\sin(3x) = 3 \sin(x) - 4 \sin^3(x)$.

The same steps on $\cos(x + 2x)$ give $\cos(3x) = 4 \cos^3(x) - 3 \cos(x)$ — also one function only, cosine.

Dividing one by the other, the same way tangent’s sum formula was built earlier, gives $\tan(3x) = (3 \tan(x) - \tan^3(x))/(1 - 3 \tan^2(x))$.

*Nothing here is a new rule. $3x$ is $x + 2x$; the sum formula and the double-angle formulas, substituted and simplified, are the whole derivation.*

Near 0 the two curves run together, which is why tripling the sine looks right for small angles. Then 3 sin x climbs to 3 while sin 3x turns back below 1. The shaded gap between them is 4 sin³x, the term the formula subtracts; at x = π/6 it is 1/2, as in the worked example.
Worked example

Given $\sin(\theta) = 1/2$, find $\sin(3 \theta)$

  1. $\sin(3 \theta) = 3 \sin(\theta) - 4 \sin^3(\theta)$
    the triple-angle formula — only sine is needed, nothing else
  2. $= 3(1/2) - 4(1/2)^3$
    substitute $\sin(\theta) = 1/2$
  3. $= 3/2 - 1/2 = 1$
    simplify
  4. $\theta = 30°$, $3 \theta = 90°$, $\sin(90°) = 1$
    check directly — $\sin(\theta) = 1/2$ is the standard value at 30 degrees, and the triple-angle answer matches the direct value exactly

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Turning a sum into a product

▲CONCEPT

The sum and difference formulas run one way: two angles combine into one. Run them the other way — add or subtract a sum formula and a difference formula — and a SUM of two ratios turns into a PRODUCT of two ratios.

Add $\sin(x+y)$ and $\sin(x-y)$: the $\cos(x)\sin(y)$ terms cancel, leaving $\sin(x+y) + \sin(x-y) = 2 \sin(x) \cos(y)$. Subtract instead and the $\sin(x)\cos(y)$ terms cancel, leaving $\sin(x+y) - \sin(x-y) = 2 \cos(x) \sin(y)$. The same two moves on the cosine sum and difference formulas give two more.

Rename the angles so they match a sum and a difference you are actually given: let $A = x+y$ and $B = x-y$, so $x = (A+B)/2$ and $y = (A-B)/2$. The four results become:

$\sin(A) + \sin(B) = 2 \sin((A+B)/2) \cos((A-B)/2)$

$\sin(A) - \sin(B) = 2 \cos((A+B)/2) \sin((A-B)/2)$

$\cos(A) + \cos(B) = 2 \cos((A+B)/2) \cos((A-B)/2)$

$\cos(A) - \cos(B) = -2 \sin((A+B)/2) \sin((A-B)/2)$

Only the last one carries a leading minus sign.

A sum of two ratios becomes a product of two ratios — useful whenever a quotient of sums needs something to cancel.

The sum wave never crosses the dashed outline 2 cos x and its mirror −2 cos x, and it is 0 wherever the outline pinches to 0, at π/2 and 3π/2. That is the product 2 sin 4x cos x: a fast wave sin 4x scaled by the slow factor 2 cos x.
Worked example

Simplify $(\sin(5x) + \sin(3x))/(\cos(5x) + \cos(3x))$

  1. $(A+B)/2 = (5x+3x)/2 = 4x$
    with $A=5x$ and $B=3x$, half the sum is the first angle the sum-to-product formulas will produce
  2. $(A-B)/2 = (5x-3x)/2 = x$
    half the difference is the second
  3. $\sin(5x) + \sin(3x) = 2 \sin(4x) \cos(x)$
    sine-plus-sine formula
  4. $\cos(5x) + \cos(3x) = 2 \cos(4x) \cos(x)$
    cosine-plus-cosine formula
  5. $(2 \sin(4x) \cos(x))/(2 \cos(4x) \cos(x))$
    substitute both products
  6. $= \tan(4x)$
    the $\cos(x)$ factor cancels top and bottom, leaving sine over cosine of $4x$
The fraction draws the same branches as tan 4x. At x = π/2, cos x = 0 makes it 0/0, so cancelling cos x holds everywhere except at that one point.

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Why everything repeats

▲CONCEPT

Go once round the circle and you are back where you started, on the very same point with the very same coordinates.

So sine and cosine repeat: $\sin(\theta + 2 \pi) = \sin(\theta)$ and $\cos(\theta + 2 \pi) = \cos(\theta)$, for every $\theta$. Their PERIOD is $2 \pi$.

Tangent repeats twice as often. Its period is $\pi$, not $2 \pi$: $\tan(\theta + \pi) = \tan(\theta)$.

The reason is worth seeing. Half a turn sends the point to the exactly opposite side of the circle, which flips the sign of BOTH coordinates. Tangent is y over x, and two sign flips cancel in a division. So tangent comes back after half a turn, while sine and cosine each need the full one.

Sine and cosine need a full turn. Tangent only needs half, because a division does not notice two sign flips.

Both brackets mark one full repeat, and the lower one is exactly half the width of the upper one. Read straight off the picture, before doing any algebra, that tangent finishes its cycle twice as often as sine and cosine do.

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Domain, range, and the six graphs

▲CONCEPT

Read the unit circle for long enough and the domain and range of every trig function become visible, not just computable.

Sine and cosine accept any angle at all — walk around the circle as many times as you like, forward or backward, and a coordinate is always waiting. Their DOMAIN is all reals. But the coordinates themselves never leave $[-1, 1]$: the circle has radius 1, so neither coordinate can exceed it. Their RANGE is $[-1, 1]$.

Tangent is different. It is $y$ over $x$, and division breaks the moment $x = 0$ — which happens at $\theta = (2n+1) \pi/2$, where the circle crosses the vertical axis. Away from those points, the ratio climbs without bound as the circle approaches them, so tangent’s RANGE is all reals even though its DOMAIN has gaps. Cotangent breaks the same way at the opposite axis crossing, $\theta = n \pi$.

Secant and cosecant invert cosine and sine. Inverting a number in $[-1, 1]$ (where it is not 0; at 0 the reciprocal is undefined) always lands outside $(-1, 1)$ — so wherever cosine or sine is defined, its reciprocal’s range is $(-\infty, -1] \cup [1, \infty)$.

The circle itself is the domain-and-range table. Where the curve goes is the range; where it cannot be evaluated at all is the missing domain.

Read straight off the curves: a dashed green line the wave never crosses marks a RANGE bound; a dashed red line marks a point where tangent is simply not defined, repeating every π. The table underneath states the same facts for all six functions.

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Solving means naming every repeat

▲CONCEPT

Because these functions repeat, an equation like $\sin(\theta) = 1/2$ never has just one solution. It has infinitely many, and a complete answer names all of them.

The three general solutions are these.

The three differ because the functions repeat differently. Tangent’s is simplest, since tangent repeats every $\pi$. Cosine’s carries a $\pm$ because $\cos(-\alpha) = \cos(\alpha)$, so a solution’s negative is also a solution. Sine’s $(-1)^n$ does the same job in a compact way, alternating the sign as n steps.

Writing “the general solution” means writing the whole family. One value found and stopped at is an incomplete answer, not a shorter one.

Find one angle, then name the whole family. The family is what was asked for.

A pair centred on zero can be named with a single plus-or-minus; a pair centred anywhere else cannot. That is the whole reason the cosine’s general solution carries one sign symbol and the sine’s has to carry an alternating one instead.
Nothing changes between the two curves except what sits inside the sine, and the count of answers doubles. That is why “solve” cannot mean “find a value”: the number of solutions in a single turn is not fixed, and it is not one. Reading the crossings off the picture also shows what the n in a general solution is counting --- the same set of crossings, once for every turn in either direction.
Worked example

Solve $\sin(\theta) = 1/2$

  1. $\sin(\pi/6) = 1/2$
    find one angle that works — $1/2$ is a standard value, so $\alpha = \pi/6$
  2. $\theta = n \pi + (-1)^n (\pi/6)$
    apply the general solution for sine, for any integer n
  3. $n=0$: $\theta = \pi/6$
    check the first value of n — $0 + \pi/6$
  4. $n=1$: $\theta = 5 \pi/6$
    check the second value of n — $(-1)^1 = -1$, so $\theta = \pi - \pi/6$
  5. $\pi/6$ and $5 \pi/6$
    both genuine solutions inside one full circle, matching what the unit circle predicts — sine is positive in Quadrants I and II, one solution lands in each; find the one easy angle first, the formula generates every other solution from it
Two crossings, one level. The second answer to this equation is fixed by the picture before any formula is applied, and the alternating sign in the general solution is only how that mirror gets written down.
Inside the shaded turn the line meets the curve twice, not once, and that is why a single equation has two answers in every turn. The first crossing is on the way up and the second on the way down. The alternating sign in the general formula is exactly that alternation, with even n picking a rising crossing and odd n a falling one. Everything to the right of the band is those same two answers moved along by a full turn.

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One picture, every idea

■RECAP
l θ r Measuring the angle: degrees and radians +,+ −,+ −,− +,− Signs, quadrant by quadrant 0 30 45 60 90 The values worth knowing cold θ −θ Turning the other way — odd and even cos sin 1 The Pythagorean identities A B Combining angles A B sin cos Tangent and cotangent of a sum x x x 3x Three times the angle + 2 × Turning a sum into a product 2π Why everything repeats Domain, range, and the six graphs θ−2π θ θ+2π Solving means naming every repeat
Four levels of cause and effect: the angle’s coordinates fix the signs and the standard values, those feed the identities, the identities combine into sums and products of angles, and periodicity is what makes solving mean naming every repeat, not finding one answer.

Every idea in this chapter starts from one picture: the unit circle, which turns an angle into a pair of coordinates.

Sine and cosine are read straight off those coordinates. RADIANS measure the angle in the way that makes arc length come out as $l = r \theta$. The QUADRANT fixes a ratio’s sign, and the PYTHAGOREAN IDENTITY fixes its size — together they pin a ratio down completely.

The SUM, DIFFERENCE and DOUBLE-ANGLE formulas build new angles’ ratios out of old ones. That is how a handful of memorised values reaches far beyond itself.

And because the picture repeats every full turn, solving an equation means naming every repeat, not just the first one you find.

One circle underneath all of it. When a step stops making sense, go back to the picture and read the coordinates.

Nothing in this drawing is new. What is new is that five separate lessons arrive on one picture, at one angle, where each can be checked against the others.

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Two traps this chapter sets

✕MISCONCEPTION

THE TRAP. $\sin(A+B) = \sin(A) + \sin(B)$. Sine distributes over addition, the way multiplication does.

THE REALITY. It does not. Sine is not linear, and no amount of wanting it to be changes that.

One test settles it. Take $A = B = 45$ degrees, so $A + B = 90$ degrees.

The left side is $\sin(90°) = 1$.

The right side is $\sin(45°) + \sin(45°) = 1/\sqrt{2} + 1/\sqrt{2}$, which is $2/\sqrt{2}$, and that is $\sqrt{2}$ — about 1.414.

So the shortcut claims $1 = \sqrt{2}$. That is not a near miss. It is wrong by more than 40 per cent.

The correct identity is the sum formula: $\sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B)$.

A single numerical check kills this one in ten seconds. Run it whenever a shortcut feels too convenient.

The shortcut is not off by a little. Fed 45 and 45 it returns about 1.41, and the graph shows there is no angle anywhere whose sine is that --- the curve never leaves the strip between minus one and one. So a single check at two angles you know cold is enough to retire the idea for good, and the real identity earns its extra terms.
✕MISCONCEPTION

THE TRAP. $\cos^2(\theta)$ means: take theta, square it, then take the cosine. That is, it means $\cos(\theta^2)$.

THE REALITY. $\cos^2(\theta)$ is shorthand for $(\cos(\theta))^2$. Take the cosine FIRST, then square what comes out.

The order is the whole of it. One squares the input. The other squares the output. They are different functions, and they give different numbers.

Try it at $\theta = \pi/3$. $\cos(\pi/3) = 1/2$, so $\cos^2(\pi/3) = 1/4$. The other reading asks for $\cos(\pi^2/9)$, the cosine of about 1.10, which is about 0.46.

The notation is admittedly poor — it puts the exponent where an argument would be. But it is standard, and every identity in this chapter assumes it. $\sin^2(\theta) + \cos^2(\theta) = 1$ is only true under the correct reading.

The exponent sits on the function name but acts on the answer. Read it as: do the cosine, then square.

These are not two spellings of one thing. Squaring the output can never produce a negative number, so the left curve stays on one side of the axis --- which is exactly what lets a sum of two squares equal 1. Squaring the angle first feeds cosine an input that grows faster and faster, so the right curve keeps crossing zero and its waves crowd up. Read both at theta equals 2 and the two answers differ in size and in sign.

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Practice set

Exercise 3.1 — Angles, degrees, radians, and arc length
  1. practice Convert 120 degrees to radian measure.
  2. practice Convert $(3 \pi)/4$ radians to degree measure.
    1. $135$
    2. $120$
    3. $150$
    4. $160$
  3. practice Find the length of an arc of a circle of radius 5 cm that subtends an angle of 60 degrees at the centre.
Answers
  1. $120 \times \pi/180 = (2 \pi)/3$
  2. $135$
  3. $60$ degrees is $\pi/3$ radians, so $l = r \theta = 5 \times \pi/3 = (5 \pi)/3$ cm
Exercise 3.1 is worth doing twice --- once with the formula, and once by putting each answer somewhere its size can be seen.
Exercise 3.2 — Quadrants, standard values, and ratios
  1. practice In which quadrant does $\theta$ lie if $\sin(\theta) > 0$ and $\cos(\theta) < 0$?
    1. Quadrant I
    2. Quadrant II
    3. Quadrant III
    4. Quadrant IV
  2. practice Find the value of $\sin(30°) + \cos(60°)$.
  3. practice If $\cos(\theta) = -3/5$ and $\theta$ lies in Quadrant III, find $\sin(\theta)$ and $\tan(\theta)$.
  4. practice Which of these equals $\sec^2(\theta) - \tan^2(\theta)$ for every $\theta$ where both are defined?
    1. $0$
    2. $1$
    3. $-1$
    4. $2 \tan^2(\theta)$
  5. practice Evaluate $\sin^2(30°) + \cos^2(60°) - \tan^2(45°)$.
Answers
  1. Quadrant II
  2. $1$
  3. $\sin(\theta) = -4/5$; $\tan(\theta) = 4/3$
  4. $1$
  5. $-1/2$
A single test angle can only ever settle a single test angle. What makes this one an identity is that the two curves keep their distance the whole way.
Exercise 3.3 — Sum, difference, and double-angle formulas
  1. practice Find the value of $\cos(75°)$ using $75 = 45 + 30$.
  2. practice If $\cos(\theta) = 12/13$ and $\theta$ lies in Quadrant I, find $\sin(2 \theta)$.
  3. practice $\cos(2A)$ equals which of these?
    1. $2 \cos^2(A) - 1$
    2. $1 - 2 \sin^2(A)$
    3. $\cos^2(A) - \sin^2(A)$
    4. All of these
  4. practice Prove that $\sin(A+B) \sin(A-B) = \sin^2(A) - \sin^2(B)$.
  5. practice Find the general solution of $\cos(\theta) = -1/2$.
  6. practice Prove that $\cos(4A) = 1 - 8 \sin^2(A) \cos^2(A)$.
  7. practice Find the exact value of $\tan(75°)$ using $75 = 45+30$.
  8. practice If $\tan(A) = 1/2$ and $\tan(B) = 1/3$, find $\cot(A+B)$.
  9. practice If $\tan(A) = 3/4$, find $\tan(2A)$.
  10. practice Find $\cos(3x)$ given $\cos(x) = 1/2$ (so $x=60°$), and check it against $\cos(180°)$.
  11. practice Write $\sin(40°)+\sin(20°)$ as a product, and evaluate it.
  12. practice Find $\cos(110°)+\cos(70°)$ using the sum-to-product formula.
Answers
  1. $\cos(75°) = (\sqrt{3}-1)/(2 \sqrt{2})$
  2. $\sin(2 \theta) = 120/169$
  3. All of these
  4. Expand both products with the sum and difference formulas; the cross terms cancel, leaving $\sin^2(A) - \sin^2(B)$.
  5. $\theta = 2 n \pi \pm (2 \pi)/3$, for any integer $n$
  6. Use $\cos(4A) = 1 - 2 \sin^2(2A)$, then substitute $\sin(2A) = 2 \sin(A) \cos(A)$.
  7. $2+\sqrt{3}$
  8. 1
  9. 24/7
  10. -1
  11. $\cos(10°) \approx 0.985$
  12. 0
Proving an identity is not confirming it once. The dashed near-miss would pass a spot check at any of the angles where the two curves happen to cross.
Miscellaneous — practice set
  1. practice If $\tan(\theta) = -4/3$ and $\theta$ lies in Quadrant II, find $\sin(\theta)$ and $\cos(\theta)$.
  2. practice Which of these functions has period $\pi$, not $2 \pi$?
    1. $\sin(\theta)$
    2. $\cos(\theta)$
    3. $\tan(\theta)$
    4. $\sin(\theta) + \cos(\theta)$
  3. practice If $\sin(\theta) = -12/13$ and $\theta$ lies in Quadrant IV, find $\cos(\theta)$ and $\cot(\theta)$.
  4. practice State the domain of $\tan(\theta)$.
  5. practice State the range of $\sec(\theta)$.
Answers
  1. $\sin(\theta) = 4/5$; $\cos(\theta) = -3/5$
  2. $\tan(\theta)$
  3. $\cos(\theta) = 5/13$; $\cot(\theta) = -5/12$
  4. $\theta \neq (2n+1)\pi/2$
  5. $\sec(\theta) \in (-\infty,-1] \cup [1,\infty)$
The distractor here is the sum, and one marked point settles it: a function with period pi would have to peak again halfway between two peaks.
A sine value is read off a horizontal line; a tangent value is read off a line through the centre. Both cut the circle twice, and both leave the quadrant to choose.

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Chapter-end problems

Chapter-end problems — graded set
  1. board-easy Convert 60° to radians.
  2. board-easy Convert $5\pi/3$ radians to degrees.
  3. board-easy Find the arc length for a circle of radius 6 and angle $\pi/3$.
  4. board-easy In which quadrant is $\sin(\theta) > 0$ and $\cos(\theta) < 0$?
  5. board-easy Evaluate $\sin(60°)$.
    1. $1/2$
    2. $\sqrt{3}/2$
    3. $1$
    4. $\sqrt{2}/2$
  6. board-easy Evaluate $\tan(45°) + \cos(0°)$.
  7. board-easy Evaluate $\cos(-30°)$.
  8. board-easy Given $\cos(\theta) = 4/5$ with theta in Quadrant I, find $\sin(\theta)$.
  9. board-easy If $\tan(\theta) = 3/4$, find $\sec^2(\theta)$.
  10. board-easy Simplify $\sin(\theta + 2 \pi)$ in terms of $\sin(\theta)$.
  11. board-easy Given $\tan(\theta) = 2$, find $\tan(\theta + \pi)$.
  12. board-easy Use the sum formula to find $\cos(60°+30°)$, and check it matches $\cos(90°)$.
  13. board-easy Is $\sin(A+B) = \sin(A)+\sin(B)$ true in general? Test it at $A=B=45°$.
  14. board-standard Convert $-150°$ to radians, and name the quadrant it lies in.
  15. board-standard A wheel of radius 0.3 m turns through 40°. Find the arc length traced by a point on its rim.
  16. board-standard Given $\cos(\theta) = -5/13$ with theta in Quadrant III, find $\sin(\theta)$ and $\tan(\theta)$.
  17. board-standard Verify that $1+\tan^2(45°) = \sec^2(45°)$.
  18. board-standard Find $\sin(15°)$ using $15 = 45-30$.
  19. board-standard Given $\sin(\theta)=3/5$ with theta in Quadrant I, find $\sin(2 \theta)$.
    1. $7/25$
    2. $24/25$
    3. $12/25$
    4. $1$
  20. board-standard Given $\sin(\theta)=1/3$, find $\cos(2 \theta)$ using the $1-2\sin^2(\theta)$ form.
  21. board-standard Solve $\cos(\theta) = -1/2$ for the general solution.
  22. board-standard Solve $\tan(\theta) = 1$ for the general solution.
    1. $\theta = n \pi + \pi/4$
    2. $\theta = 2 n \pi + \pi/4$
    3. $\theta = n \pi - \pi/4$
    4. $\theta = n \pi/4$
  23. board-standard Simplify $\sin(\theta) \cos(\theta) (\tan(\theta) + \cot(\theta))$.
  24. board-standard Given $\cos(\theta) = 1/2$, find $\cos^2(\theta)$ — not $\cos(\theta^2)$.
  25. board-standard Given $\sin(\theta) = 5/13$ with theta in Quadrant II, find $\cos(\theta)$ and $\cos(2 \theta)$.
  26. JEE Evaluate $\sin(105°)$ exactly.
    1. $(\sqrt{6}+\sqrt{2})/4$
    2. $(\sqrt{6}-\sqrt{2})/4$
    3. $(\sqrt{3}+1)/4$
    4. $\sqrt{2}/2$
  27. JEE If $\cos(\theta) = -3/5$ with theta in Quadrant II, find $\sin(2 \theta)$ and $\cos(2 \theta)$.
  28. JEE Prove that $(1-\cos(2 \theta))/(\sin(2 \theta)) = \tan(\theta)$.
  29. JEE Solve $2\sin^2(\theta) - \sin(\theta) - 1 = 0$ for the general solution.
  30. JEE Given $\tan(\theta) = -4/3$ with theta in Quadrant II, find $\sin(\theta)$ and $\cos(\theta)$.
  31. JEE If $A+B=45°$, show that $(1+\tan(A))(1+\tan(B)) = 2$.
  32. JEE Find the general solution of $\cos(\theta) + \cos(2 \theta) = 0$.
  33. JEE Simplify $\sin^4(\theta) - \cos^4(\theta)$ to a single term in theta.
  34. JEE Solve $\tan^2(\theta) = 3$ for $0 \leq \theta < 2 \pi$.
  35. JEE Given $\sin(A) = 3/5$, $\cos(B) = 12/13$, with A, B both in Quadrant I, find $\sin(A+B)$.
Answers
  1. $\pi/3$
  2. 300°
  3. $2 \pi$
  4. Quadrant II
  5. $\sqrt{3}/2$
  6. 2
  7. $\sqrt{3}/2$
  8. 3/5
  9. 25/16
  10. $\sin(\theta)$
  11. 2
  12. 0
  13. false — sine is not linear
  14. $-5\pi/6$, Quadrant III
  15. $\pi/15 \approx 0.21$ m
  16. $\sin(\theta) = -12/13$, $\tan(\theta) = 12/5$
  17. confirmed — both sides equal 2
  18. $(\sqrt{6}-\sqrt{2})/4$
  19. 24/25
  20. 7/9
  21. $\theta = 2 n \pi \pm 2\pi/3$
  22. $\theta = n \pi + \pi/4$
  23. 1
  24. 1/4
  25. $\cos(\theta) = -12/13$, $\cos(2 \theta) = 119/169$
  26. $(\sqrt{6}+\sqrt{2})/4$
  27. $\sin(2 \theta) = -24/25$, $\cos(2 \theta) = -7/25$
  28. identity proved
  29. $\theta = n \pi + (-1)^n (-\pi/6)$ or $\theta = 2 n \pi+\pi/2$
  30. $\sin(\theta) = 4/5$, $\cos(\theta) = -3/5$
  31. identity proved, equals 2
  32. $\theta = 2 n \pi \pm \pi/3$ or $\theta=(2n+1) \pi$
  33. $-\cos(2 \theta)$
  34. $\theta = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$
  35. 56/65
Two points, one curve: the sign inside cosine’s formula does not shade the answer’s size, it selects which angle the answer belongs to.
JEE-tier practice — from the item bank
  1. Evaluate $\cos(150°)$.
    1. $\sqrt{3}/2$
    2. $-\sqrt{3}/2$
    3. $-1/2$
    4. $-2/\sqrt{3}$
    Check your answer
    1. $\sqrt{3}/2$ — 150 degrees has reference angle 30 degrees, so the SIZE is cos(30 degrees) = $\sqrt{3}/2$ — but Quadrant II makes cosine negative, so the value itself must be negative.
    2. ✓ $-\sqrt{3}/2$ — cos(150°) = −cos(30°) = $−\sqrt{3}/2$ — Quadrant II makes cosine negative, and the reference angle 30 degrees supplies the size.
    3. $-1/2$ — 1/2 is sin(30 degrees), not cos(30 degrees) — the reference angle’s cosine is $\sqrt{3}/2.$
    4. $-2/\sqrt{3}$ — $−2/\sqrt{3}$ is sec(150 degrees), the reciprocal of cos(150 degrees) — a different ratio of the same angle.
  2. Given $\sin(\theta) = -1/2$ with theta in Quadrant III, find $\cos(\theta)$.
    1. $-2/\sqrt{3}$
    2. $\sqrt{3}/2$
    3. $-\sqrt{3}/2$
    4. $-1/2$
    Check your answer
    1. $-2/\sqrt{3}$ — $−2/\sqrt{3}$ is $\sec(\theta)$, the reciprocal of $\cos(\theta)$ — a different ratio built from the same angle.
    2. $\sqrt{3}/2$ — $\cos^2(\theta)$ does work out to 3/4, so the SIZE $\sqrt{3}/2$ is right — but Quadrant III makes cosine negative, so the value itself must be negative.
    3. ✓ $-\sqrt{3}/2$ — $\cos^2(\theta)$ = 1 − 1/4 = 3/4, so $\cos(\theta)$ = plus.minus $\sqrt{3}/2$ — Quadrant III has cosine negative, so $\cos(\theta)$ = $−\sqrt{3}/2.$
    4. $-1/2$ — This is sine’s own value, not a cosine computed from the Pythagorean identity.
  3. Solve $\sin(\theta) = -1/2$ for the general solution.
    1. $\theta = n \pi + (-1)^n (-\pi/6)$
    2. $\theta = n \pi + (-1)^n (\pi/6)$
    3. $\theta = 2 n \pi \pm \pi/6$
    4. $\theta = n \pi + \pi/6$
    Check your answer
    1. ✓ $\theta = n \pi + (-1)^n (-\pi/6)$ — $\sin(−\pi/6)$ = −1/2, so alpha = $−\pi/6$ and the general solution is theta = n pi + $(−1)^n$ $(−\pi/6)$.
    2. $\theta = n \pi + (-1)^n (\pi/6)$ — The formula for $\sin(\theta)$ = sin(alpha) is theta = n pi + $(−1)^n$ alpha. Here alpha = $−\pi/6$ itself (since $\sin(−\pi/6)$ = −1/2), so the formula must carry that negative sign.
    3. $\theta = 2 n \pi \pm \pi/6$ — That is cosine’s general-solution shape, 2 n pi plus.minus alpha — a different equation gets a different formula.
    4. $\theta = n \pi + \pi/6$ — That is tangent’s general-solution shape, n pi + alpha — sine’s own formula carries the $(−1)^n$ factor, tangent’s does not.
  4. Given $\sin(A) = 7/25$ and $\cos(B) = 4/5$, with A, B both in Quadrant I, find $\cos(A-B)$.
    1. $96/125$
    2. $44/125$
    3. $117/125$
    4. $-117/125$
    Check your answer
    1. $96/125$ — cos(A−B) = cos(A)cos(B) + sin(A)sin(B) has TWO terms added together — this option keeps only the first one.
    2. $44/125$ — The formula pairs cos with cos and sin with sin, not cos(A) with sin(B).
    3. ✓ $117/125$ — cos(A) = 24/25 and sin(B) = 3/5 from the Pythagorean identity; cos(A−B) = (24/25)(4/5) + (7/25)(3/5) = 96/125 + 21/125 = 117/125.
    4. $-117/125$ — cos(A−B) ADDS the two products; only cos(A+B) subtracts them, which would flip this sign.
  5. Given $\cos(\theta) = 7/25$ with theta in Quadrant I, find $\cos(2 \theta)$.
    1. $527/625$
    2. $-337/625$
    3. $49/625$
    4. $-527/625$
    Check your answer
    1. $527/625$ — $2(7/25)^2$ − 1 does equal −527/625, not +527/625 — the final subtraction of 1 must be carried through.
    2. $-337/625$ — The double-angle formula is $2\cos^2(\theta)$ − 1, not $\cos^2(\theta)$ − 1 — the squared term is doubled before subtracting.
    3. $49/625$ — $\cos^2(\theta)$ = 49/625 is only an intermediate value — $\cos(2 \theta)$ still needs the double, then the minus 1.
    4. ✓ $-527/625$ — $\cos(2 \theta)$ = $2\cos^2(\theta)$ − 1 = 2(49/625) − 1 = 98/625 − 625/625 = −527/625.
  6. Given $\tan(\theta) = 5/12$ with theta in Quadrant III, find $\sin(\theta)$.
    1. $5/13$
    2. $-12/13$
    3. $12/13$
    4. $-5/13$
    Check your answer
    1. $5/13$ — 5/13 is the correct SIZE, but Quadrant III makes sine negative — the value itself must be negative.
    2. $-12/13$ — −12/13 is $\cos(\theta)$, not $\sin(\theta)$ — both are negative in Quadrant III, but they are different ratios.
    3. $12/13$ — 12/13 is cosine’s size, and this answer also keeps sine’s expected sign — neither ratio’s parts should be mixed like this.
    4. ✓ $-5/13$ — $\sec^2(\theta)$ = 1+25/144 = 169/144, so $\cos(\theta)$ = −12/13 (Quadrant III). Then $\sin(\theta)$ = $\tan(\theta)$ times $\cos(\theta)$ = (5/12)(−12/13) = −5/13.
  7. Solve $\sqrt{3} \tan(\theta) = 1$ for the general solution.
    1. $\theta = 2 n \pi + \pi/6$
    2. $\theta = n \pi + \pi/6$
    3. $\theta = n \pi - \pi/6$
    4. $\theta = n \pi + \pi/3$
    Check your answer
    1. $\theta = 2 n \pi + \pi/6$ — Tangent’s general solution is theta = n pi + alpha, with a plain n pi — 2n pi belongs to cosine’s formula, not tangent’s.
    2. ✓ $\theta = n \pi + \pi/6$ — $\tan(\theta)$ = $1/\sqrt{3}$ = $\tan(\pi/6)$, and tangent’s general solution is theta = n pi + alpha, so theta = n pi + $\pi/6.$
    3. $\theta = n \pi - \pi/6$ — $\tan(\theta)$ = tan(alpha) gives theta = n pi + alpha, with a PLUS sign, matching alpha = $\pi/6$ directly.
    4. $\theta = n \pi + \pi/3$ — $\tan(\pi/3)$ = $\sqrt{3}$, not $1/\sqrt{3}$ — the standard angle whose tangent is $1/\sqrt{3}$ is $\pi/6.$
  8. Express $(\sin(\theta)+\cos(\theta))^2$ in terms of $\sin(2 \theta)$.
    1. $\sin(2 \theta)$
    2. $1-\sin(2 \theta)$
    3. $1+\cos(2 \theta)$
    4. $1+\sin(2 \theta)$
    Check your answer
    1. $\sin(2 \theta)$ — Expanding the square gives $\sin^2(\theta)$ + $2\sin(\theta)\cos(\theta)$ + $\cos^2(\theta)$ — the constant 1 (from the Pythagorean identity) is a real term, not to be dropped.
    2. $1-\sin(2 \theta)$ — $2\sin(\theta)\cos(\theta)$ is $\sin(2 \theta)$ with a PLUS sign, from the expansion’s middle term — nothing flips it negative here.
    3. $1+\cos(2 \theta)$ — $\cos(2 \theta)$ is a different double-angle expression; the cross term $2\sin(\theta)\cos(\theta)$ matches $\sin(2 \theta)$ specifically.
    4. ✓ $1+\sin(2 \theta)$ — $(\sin(\theta)+\cos(\theta))^2$ = $\sin^2(\theta)+\cos^2(\theta)$ + $2\sin(\theta)\cos(\theta)$ = 1 + $\sin(2 \theta)$, using the Pythagorean identity and the double-angle formula for sine.
  9. Evaluate $\sin(75°)\cos(15°) - \cos(75°)\sin(15°)$.
    1. $\sqrt{3}/2$
    2. $1$
    3. $1/2$
    4. $\sqrt{2}/2$
    Check your answer
    1. ✓ $\sqrt{3}/2$ — sin(A)cos(B) − cos(A)sin(B) = sin(A−B), so this is sin(75−15) degrees = sin(60 degrees) = $\sqrt{3}/2$ — no separate values need computing.
    2. $1$ — This expression is sin(A)cos(B) − cos(A)sin(B), which collapses to sin(A−B), not sin(A+B) — 75−15 = 60 degrees, not 90.
    3. $1/2$ — The angle that forms is 75−15 = 60 degrees, whose sine is $\sqrt{3}/2$, not sin(30 degrees).
    4. $\sqrt{2}/2$ — 45 degrees is not the angle this expression collapses to — 75 minus 15 is 60 degrees.
  10. If $A=60°$ and $B=30°$, evaluate $\cos(A+B)\cos(A-B)$.
    1. $1$
    2. $1/4$
    3. $-1/4$
    4. $0$
    Check your answer
    1. $1$ — The two cosines have to be computed from the actual angles A+B=90 degrees and A−B=30 degrees, not assumed.
    2. $1/4$ — cos^2(60 degrees) = 1/4 is a different quantity — the question asks for cos(A+B)cos(A−B), not $\cos^2(A)$.
    3. $-1/4$ — Neither the value nor its sign matches cos(A+B)cos(A−B) here — cos(90 degrees) is 0, which makes the whole product 0.
    4. ✓ $0$ — A+B=90 degrees and A−B=30 degrees, so cos(A+B)cos(A−B) = cos(90 degrees) times cos(30 degrees) = 0 times $(\sqrt{3}/2)$ = 0.
  11. Find the smallest positive theta with $0° < \theta < 180°$ such that $\tan(\theta) = \tan(200°)$.
    1. $160°$
    2. $200°$
    3. $110°$
    4. $20°$
    Check your answer
    1. $160°$ — 360 degrees is sine and cosine’s period, not tangent’s — tangent repeats every 180 degrees.
    2. $200°$ — 200 degrees itself is outside the requested range (0,180) — the periodic equivalent inside that range is what is asked for.
    3. $110°$ — 90 degrees is not tangent’s period — subtracting it does not give an angle with the same tangent value.
    4. ✓ $20°$ — Tangent repeats every 180 degrees, so tan(200 degrees) = tan(200−180 degrees) = tan(20 degrees), and 20 degrees already lies in (0,180).
  12. If $\sec(\theta) < 0$ and $\tan(\theta) > 0$, which quadrant is theta in?
    1. Quadrant 1
    2. Quadrant 2
    3. Quadrant 3
    4. Quadrant 4
    Check your answer
    1. Quadrant 1 — In Quadrant 1 every ratio is positive, so $\sec(\theta)$ < 0 already rules this quadrant out.
    2. Quadrant 2 — Secant is negative in Quadrant 2, matching the first condition — but tangent is also negative there, failing the second condition.
    3. ✓ Quadrant 3 — Secant negative means cosine negative — Quadrant 2 or 3. Tangent positive means sine and cosine share a sign — Quadrant 1 or 3. Only Quadrant 3 satisfies both.
    4. Quadrant 4 — Cosine (and so secant) is positive in Quadrant 4, failing $\sec(\theta)$ < 0 immediately.
  13. Which of these is NOT a valid expression for $\cos(2 \theta)$?
    1. $\cos^2(\theta) - \sin^2(\theta)$
    2. $2\cos^2(\theta) - 1$
    3. $1 - 2\sin^2(\theta)$
    4. $2\sin(\theta)\cos(\theta)$
    Check your answer
    1. $\cos^2(\theta) - \sin^2(\theta)$ — $\cos^2(\theta)$ − $\sin^2(\theta)$ is the direct double-angle definition — a valid, genuine form of $\cos(2 \theta)$.
    2. $2\cos^2(\theta) - 1$ — $2\cos^2(\theta)$ − 1 comes from substituting $\sin^2(\theta)$ = 1 − $\cos^2(\theta)$ — a valid form.
    3. $1 - 2\sin^2(\theta)$ — 1 − $2\sin^2(\theta)$ comes from substituting $\cos^2(\theta)$ = 1 − $\sin^2(\theta)$ — also a valid form.
    4. ✓ $2\sin(\theta)\cos(\theta)$ — $2\sin(\theta)\cos(\theta)$ is the double-angle formula for $\sin(2 \theta)$, not $\cos(2 \theta)$ — it is the one option that names the wrong function’s identity.
  14. Evaluate $(\tan(60°)-\tan(30°))/(1+\tan(60°)\tan(30°))$.
    1. $\sqrt{3}$
    2. $-\sqrt{3}/3$
    3. $1/2$
    4. $\sqrt{3}/3$
    Check your answer
    1. $\sqrt{3}$ — This ratio has the shape of the tangent DIFFERENCE formula, tan(60−30), not tan(60) alone.
    2. $-\sqrt{3}/3$ — 60−30 = 30 degrees, and tan(30 degrees) is positive — nothing in the difference formula introduces a minus sign here.
    3. $1/2$ — 1/2 is sin(30 degrees), not tan(30 degrees) — the two are different ratios of the same angle.
    4. ✓ $\sqrt{3}/3$ — This is the tangent difference formula’s shape, giving tan(60 degrees − 30 degrees) = tan(30 degrees) = $\sqrt{3}/3.$
  15. Is $\cos^2(\theta) = \cos(\theta^2)$ true for every theta?
    1. True — squaring the angle and squaring cosine’s output are the same operation, just done in a different order
    2. False — $\cos^2(\theta)$ means $(\cos(\theta))^2$, not $\cos(\theta^2)$
    3. True, but only when theta is in Quadrant I
    4. False, because cosine can never be squared, by definition
    Check your answer
    1. True — squaring the angle and squaring cosine’s output are the same operation, just done in a different order — $\cos^2(\theta)$ squares the OUTPUT of cosine; $\cos(\theta^2)$ squares the ANGLE first, then takes cosine — these are different operations done in a different order, and generally give different numbers.
    2. ✓ False — $\cos^2(\theta)$ means $(\cos(\theta))^2$, not $\cos(\theta^2)$ — $\cos^2(\theta)$ is shorthand for $(\cos(\theta))^2$ — square the OUTPUT. $\cos(\theta^2)$ squares the ANGLE first. The two operations happen in a different order and are not equal in general.
    3. True, but only when theta is in Quadrant I — The two sides genuinely differ, but not because of which quadrant theta sits in — try any theta and the two computations diverge.
    4. False, because cosine can never be squared, by definition — $\cos^2(\theta)$, meaning $(\cos(\theta))^2$, is a perfectly ordinary, valid quantity — the statement is false for a notational reason, not because squaring cosine is somehow forbidden.
  16. Convert $5\pi/4$ radians to degrees, and name the quadrant it lies in.
    1. $200°$, Quadrant III
    2. $225°$, Quadrant IV
    3. $225°$, Quadrant II
    4. $225°$, Quadrant III
    Check your answer
    1. $200°$, Quadrant III — $(5 \pi/4)$ times $(180/\pi)$ = 5 times 45 = 225, not 200 — the multiplication itself is off.
    2. $225°$, Quadrant IV — 225 degrees lies between 180 and 270, which is Quadrant III, not Quadrant IV (270-360).
    3. $225°$, Quadrant II — 225 degrees lies between 180 and 270, which is Quadrant III, not Quadrant II (90-180).
    4. ✓ $225°$, Quadrant III — $(5 \pi/4)$ times $(180/\pi)$ = 225 degrees, and 225 degrees is between 180 and 270, so it is in Quadrant III.
  17. Evaluate $\sin(\pi/2 - \pi/6)$.
    1. $\sqrt{3}/2$
    2. $1/2$
    3. $1$
    4. $\sqrt{2}/2$
    Check your answer
    1. ✓ $\sqrt{3}/2$ — $\pi/2$ − $\pi/6$ = $\pi/3$, and $\sin(\pi/3)$ = $\sqrt{3}/2$, a standard angle value.
    2. $1/2$ — $\pi/2$ − $\pi/6$ is $\pi/3$, not $\pi/6$ — the subtraction has to be done first.
    3. $1$ — The $\pi/6$ does not vanish — $\pi/2$ − $\pi/6$ works out to $\pi/3$, a specific angle to evaluate.
    4. $\sqrt{2}/2$ — $\sqrt{2}/2$ is the 45-degree value; $\pi/2$ − $\pi/6$ = $\pi/3$ (60 degrees), a different standard angle.
  18. Given $\sin(\theta) = 4/5$ with theta in Quadrant II, find $\tan(2 \theta)$.
    1. $-24/25$
    2. $-7/24$
    3. $7/24$
    4. $24/7$
    Check your answer
    1. $-24/25$ — $\sin(2 \theta)$ = −24/25 is only half the computation — $\tan(2 \theta)$ still needs dividing by $\cos(2 \theta)$.
    2. $-7/24$ — This is $\cos(2 \theta)/\sin(2 \theta)$ with the wrong sign, not $\sin(2 \theta)/\cos(2 \theta)$.
    3. $7/24$ — This inverts $\sin(2 \theta)/\cos(2 \theta)$ — $\tan(2 \theta)$ is sine over cosine, not cosine over sine.
    4. ✓ $24/7$ — $\cos(\theta)$ = −3/5 (Quadrant II). $\sin(2 \theta)$ = 2(4/5)(−3/5) = −24/25; $\cos(2 \theta)$ = $2(−3/5)^2−1$ = −7/25. $\tan(2 \theta)$ = (−24/25)/(−7/25) = 24/7.
  19. Solve $2\cos^2(\theta) - 1 = 0$ for the general solution.
    1. $\theta = n \pi + \pi/4$
    2. $\theta = 2 n \pi \pm \pi/4$
    3. $\theta = n \pi/4$
    4. $\theta = (2n+1) \pi/4$
    Check your answer
    1. $\theta = n \pi + \pi/4$ — n pi + $\pi/4$ is tangent’s general-solution shape — this equation is really about $\cos(2 \theta)$, which needs cosine’s own formula applied to $2\theta$.
    2. $\theta = 2 n \pi \pm \pi/4$ — 2 n pi plus.minus $\pi/4$ would solve $\cos(\theta)$ = $\cos(\pi/4)$ directly — but the equation here is $2\cos^2(\theta)−1=0$, which is $\cos(2\theta)=0$ in disguise.
    3. $\theta = n \pi/4$ — theta = n $\pi/4$ includes extra values (like theta = pi/2) that do not actually satisfy $2\cos^2(\theta)−1=0.$
    4. ✓ $\theta = (2n+1) \pi/4$ — $2\cos^2(\theta)−1$ is $\cos(2 \theta)$ by the double-angle formula, so the equation is $\cos(2 \theta)$ = 0, giving $2\theta$ = $(2n+1)\pi/2$, so theta = $(2n+1)\pi/4.$
  20. Given $\csc(\theta) = -2$ with theta in Quadrant IV, find $\cot(\theta)$.
    1. $\sqrt{3}$
    2. $1/\sqrt{3}$
    3. $-1/\sqrt{3}$
    4. $-\sqrt{3}$
    Check your answer
    1. $\sqrt{3}$ — The size $\sqrt{3}$ is right, but Quadrant IV has cosine positive and sine negative, so their ratio $\cot(\theta)$ is negative, not positive.
    2. $1/\sqrt{3}$ — This is $\tan(\theta)$, the reciprocal of $\cot(\theta)$ — and with the wrong sign besides.
    3. $-1/\sqrt{3}$ — This is $\tan(\theta)$, not $\cot(\theta)$ — the ratio is inverted, even though the sign here matches.
    4. ✓ $-\sqrt{3}$ — $\csc(\theta)=−2$ means $\sin(\theta)=−1/2$; Quadrant IV gives $\cos(\theta)=\sqrt{3}/2.$ $\cot(\theta)=\cos(\theta)/\sin(\theta)$ = $(\sqrt{3}/2)/(−1/2)$ = $−\sqrt{3}$.
Written down, the three general solutions look like variations on one formula. Laid out as the angles they actually name, they are three different patterns --- which is why picking the wrong one is not a small slip.
Five separate items in this set are the same square root done five times. Recognising which of the two triangles is in play is the shortcut; the quadrant then supplies both signs.

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