VertexMath · MHF4U · Free worksheet

Trigonometric Identities Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 50 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Quick reference

Reciprocal and quotient: $\csc x=\dfrac{1}{\sin x}$ , $\sec x=\dfrac{1}{\cos x}$ , $\cot x=\dfrac{\cos x}{\sin x}$ , $\tan x=\dfrac{\sin x}{\cos x}$ . Cofunction: $\sin\left(\frac{\pi}{2}-x\right)=\cos x$ .
Pythagorean: $\sin^2x+\cos^2x=1$ and $1+\tan^2x=\sec^2x$ . A square root gives $\pm$ : the quadrant picks the sign.
Compound angle: $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ , $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ , $\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ .
Double angle: $\sin 2A=2\sin A\cos A$ , $\cos 2A=\cos^2A-\sin^2A=1-2\sin^2A=2\cos^2A-1$ , $\tan 2A=\dfrac{2\tan A}{1-\tan^2A}$ .
Proofs: work on one side at a time, never on the equation you are proving. And $\cos(A-B)\ne\cos A-\cos B$ , $\sin 2A\ne 2\sin A$ .

Show your work on separate paper. Show your work in the dot-grid space.

Core Skills (K)

Reciprocal & Quotient · Pythagorean · Compound Angle · Double Angle · Equations

K1 Reciprocal, Quotient & Cofunction

Simplify each expression to a single trig ratio or a constant.

a)
$\dfrac{\sec x}{\csc x}$
b)
$\cot x\,\sec x\,\sin x$
c)
$\tan\left(\dfrac{\pi}{2}-x\right)\sin x$

Stuck? Peek at a hint

Rewrite every ratio in terms of $\sin x$ and $\cos x$ before cancelling. For (c), start with a cofunction identity.

K2 Pythagorean Identity

For (a) and (b), $\cos\theta=-\dfrac{7}{25}$ and $\pi<\theta<\dfrac{3\pi}{2}$ .

a)
Determine $\sin\theta$ exactly.
b)
Determine $\tan\theta$ and $\csc\theta$ exactly.
c)
Simplify $\dfrac{\sec^2x-1}{\sec^2x}$ to a single term.

Stuck? Peek at a hint

Use $\sin^2\theta=1-\cos^2\theta$ , then let the quadrant choose the sign. For (c), replace $\sec^2x-1$ with one term.

K3 Compound Angles

Angles $A$ and $B$ have the exact values in the table.

$\text{angle}$	$A$	$B$
$\sin$	$\dfrac{8}{17}$	$\dfrac{3}{5}$
$\cos$	$\dfrac{15}{17}$	$-\dfrac{4}{5}$

a)
Determine $\sin(A+B)$ exactly.
b)
Determine $\cos(A-B)$ exactly.
c)
Evaluate $\tan\dfrac{5\pi}{12}$ exactly.

Stuck? Peek at a hint

Write the formula first, then substitute each value in brackets. For (c), write $\frac{5\pi}{12}$ as a sum of two special angles.

K4 Double Angles

For (a) and (b), $\sin A=\dfrac{2}{3}$ and $\dfrac{\pi}{2}<A<\pi$ .

a)
Determine $\sin 2A$ exactly.
b)
Determine $\cos 2A$ and $\tan 2A$ exactly.
c)
Write each as a single trig ratio, then evaluate exactly: $2\cos^2\dfrac{\pi}{8}-1$ and $\dfrac{2\tan 15^\circ}{1-\tan^2 15^\circ}$ .

Stuck? Peek at a hint

Find $\cos A$ first; the quadrant sets its sign. For (c), match each expression to one form of a double angle formula.

K5 Equations with Identities

Solve for $0\le x\le 2\pi$ . Give exact values.

a)
$2\sin^2x+3\cos x-3=0$
b)
$\sin 2x=\cos x$
c)
$\cos 2x=3\sin x-1$

Stuck? Peek at a hint

Use an identity to write each equation in a single trig ratio of $x$ . Then move every term to one side and factor.

Real World (A)

Compound Angles · Double Angles · Equations in Context

A1 Mural View

A mural on a Toronto building starts 3 m above your eye level and ends 8 m above it. You stand $x$ metres from the wall. The angles of elevation to the bottom and top of the mural are $\alpha$ and $\beta$ , so the mural fills an angle $\theta=\beta-\alpha$ of your view.

a)
Show that $\tan\theta=\dfrac{5x}{x^2+24}$ .
b)
Find the two distances where $\tan\theta=\frac{1}{2}$ , and the angle $\theta$ to the nearest tenth of a degree.
c)
A photographer wants the mural to fill a $45^\circ$ angle of view. How far from the wall should they stand?

Stuck? Peek at a hint

Write $\tan\alpha$ and $\tan\beta$ in terms of $x$ , then use the formula for $\tan(\beta-\alpha)$ .

A2 Ladder Check

A 6 m ladder leans against a vertical wall and makes an angle $\theta$ with the level ground, $0^\circ<\theta<90^\circ$ . The height of its top plus the distance of its foot from the wall is 8 m.

a)
Show that $\sin 2\theta=\dfrac{7}{9}$ .
b)
Find both possible angles $\theta$ to the nearest tenth of a degree, and the height of the top of the ladder for each, to the nearest centimetre.
c)
A safety guideline says a ladder should make an angle between $70^\circ$ and $80^\circ$ with the ground. Can this ladder meet the 8 m condition safely? Find the exact value of height plus distance at $75^\circ$ .

Stuck? Peek at a hint

Write the height and the distance in terms of $\theta$ , then square both sides of the equation.

A3 Indoor Kick

A soccer player kicks a ball from the ground at 20 m/s, at an angle $\theta$ above the horizontal. Ignoring air resistance, it lands $d(\theta)=\dfrac{800\sin\theta\cos\theta}{9.8}$ metres away on level ground.

a)
Write $d(\theta)$ using a single sine function. Find the angle that gives the longest kick, and that distance to the nearest tenth of a metre.
b)
Find both angles that land the ball 30 m away, to the nearest tenth of a degree.
c)
The game is in an indoor fieldhouse with a 15 m roof. The ball's maximum height is $h=\dfrac{400\sin^2\theta}{19.6}$ metres. Which angle from (b) can the player use?

Stuck? Peek at a hint

Use $2\sin\theta\cos\theta=\sin 2\theta$ . In (b), $2\theta$ can be any angle from $0^\circ$ to $180^\circ$ .

Think It Through (T)

Error Analysis · Building Identities

T1 Spot the Flaw

A student claims that $\sin x-\cos x=\sqrt{1-\sin 2x}$ is an identity:
"Square both sides: $\sin^2x-2\sin x\cos x+\cos^2x=1-\sin 2x$ . The left side equals $1-\sin 2x$ , the same as the right side. Therefore the identity is true."
The graph shows $y=\sin x-\cos x$ (LS) and $y=\sqrt{1-\sin 2x}$ (RS) for $0\le x\le2\pi$ . Vertical grid lines are $\frac{\pi}{2}$ apart and horizontal grid lines are $0.5$ apart.

a)
Evaluate both sides at $x=0$ . What does this show about the claim?
b)
State the interval in $[0,2\pi]$ on which the two sides are equal. Give exact endpoints.
c)
Identify the flaw in the student's method. Then write the identity that the squaring step does prove.

Stuck? Peek at a hint

One value of $x$ where the sides differ disproves an identity. For (b), find where $\sin x-\cos x\ge0$ .

T2 Build It Backwards

Find the constants. Each identity must hold for every real number $x$ .

a)
Find $a$ and $b$ such that $\sin^2x=a+b\cos 2x$ .
b)
Find $R>0$ and $0<\alpha<\dfrac{\pi}{2}$ such that $\sin x+\sqrt{3}\cos x=R\sin(x+\alpha)$ .
c)
A student says the largest value of $\sin x+\sqrt{3}\cos x$ is $1+\sqrt{3}$ , since $\sin x\le1$ and $\sqrt{3}\cos x\le\sqrt{3}$ . Use (b) to explain why the student is wrong. Find the largest value and the smallest positive $x$ where it occurs.

Stuck? Peek at a hint

Expand $R\sin(x+\alpha)$ with the compound angle formula and match the coefficients of $\sin x$ and $\cos x$ .

Boss Level

Optional challenge

Telescoping Cosines

Let $P=\cos20^\circ\cos40^\circ\cos80^\circ$ .

a)
Show that $8\sin20^\circ\cdot P=\sin160^\circ$ .
b)
Find the exact value of $P$ .
c)
Find the exact value of $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{4\pi}{7}$ .

Stuck? Peek at a hint

Use $2\sin\theta\cos\theta=\sin 2\theta$ again and again. Then compare $\sin160^\circ$ with $\sin20^\circ$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) $\tan x$
b) $1$
c) $\cos x$

a) $\sin\theta=-\dfrac{24}{25}$
b) $\tan\theta=\dfrac{24}{7}$ , $\csc\theta=-\dfrac{25}{24}$
c) $\sin^2x$

a) $\dfrac{13}{85}$
b) $-\dfrac{36}{85}$
c) $2+\sqrt{3}$

a) $-\dfrac{4\sqrt{5}}{9}$
b) $\cos 2A=\dfrac{1}{9}$ , $\tan 2A=-4\sqrt{5}$
c) $\cos\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$ ; $\tan 30^\circ=\dfrac{\sqrt{3}}{3}$

a) $x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3},\ 2\pi$
b) $x=\dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$
c) $x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}$

Real World

a) $\tan\theta=\dfrac{\frac{8}{x}-\frac{3}{x}}{1+\frac{8}{x}\cdot\frac{3}{x}}=\dfrac{5x}{x^2+24}$
b) $x=4$ m or $x=6$ m; $\theta\approx26.6^\circ$ at both
c) Nowhere: $\tan\theta=1$ gives $x^2-5x+24=0$ , with discriminant $-71<0$ . The mural never fills $45^\circ$ from eye level.

a) $6\sin\theta+6\cos\theta=8$ , so $(\sin\theta+\cos\theta)^2=\frac{16}{9}$ , $1+\sin 2\theta=\frac{16}{9}$ and $\sin 2\theta=\frac{7}{9}$
b) $\theta\approx25.5^\circ$ : height $\approx2.59$ m; $\theta\approx64.5^\circ$ : height $\approx5.41$ m
c) No: $25.5^\circ$ and $64.5^\circ$ are both outside $[70^\circ,80^\circ]$ . At $75^\circ$ : $3\sqrt{6}\approx7.35$ m.

a) $d(\theta)=\dfrac{400\sin 2\theta}{9.8}$ ; $\theta=45^\circ$ ; $40.8$ m
b) $\theta\approx23.7^\circ$ or $\theta\approx66.3^\circ$
c) Only $23.7^\circ$ ( $h\approx3.3$ m). At $66.3^\circ$ , $h\approx17.1$ m: the ball hits the roof.

Think It Through

a) LS $=-1$ , RS $=1$ : the claim is not an identity.
b) $\left[\dfrac{\pi}{4},\dfrac{5\pi}{4}\right]$
c) The student started from the claim and squared both sides; squaring can turn a false equation ( $-1=1$ ) into a true one. It proves $(\sin x-\cos x)^2=1-\sin 2x$ , i.e. $|\sin x-\cos x|=\sqrt{1-\sin 2x}$ .

a) $a=\dfrac{1}{2}$ , $b=-\dfrac{1}{2}$
b) $R=2$ , $\alpha=\dfrac{\pi}{3}$
c) The two terms never peak together ( $\sin x=1$ forces $\cos x=0$ ). Largest value $2$ , at $x=\dfrac{\pi}{6}$ .

Boss Level

Boss

a) $8\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ=4\sin40^\circ\cos40^\circ\cos80^\circ=2\sin80^\circ\cos80^\circ=\sin160^\circ$
b) $P=\dfrac{1}{8}$
c) $-\dfrac{1}{8}$

Free cheat sheet

Trigonometric Identities Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

One printable page with every rule.

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Most testedGiven $\sin A=3/5$ with $A$ in Quadrant II, find $\sin2A$ and $\cos2A$ . (card 4)

1Reciprocal, quotient, cofunction

$\csc x=\dfrac1{\sin x}\quad\sec x=\dfrac1{\cos x}\quad\cot x=\dfrac{\cos x}{\sin x}$

$\sin(\pi/2-x)=\cos x$

$\tan(\pi/2-x)=\cot x$

Trap $\sec$ goes with $\cos$ , $\csc$ with $\sin$ : $\sec x/\csc x=\sin x/\cos x=\tan x$ .

2Pythagorean identities

$\sin^2x+\cos^2x=1\qquad1+\tan^2x=\sec^2x$

UseA square root gives $\pm$ : the quadrant picks the sign.

Example $\cos\theta=-7/25$ in QIII: $\sin\theta=-\sqrt{1-49/625}=-24/25$ .

Trap $\sqrt{1-\cos^2\theta}=|\sin\theta|$ , not $1-\cos\theta$ .

3Compound angles

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$

$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$

$\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

Example $\cos75^\circ=\cos(45^\circ+30^\circ)=(\sqrt6-\sqrt2)/4$ .

Trap $\cos(A-B)\ne\cos A-\cos B$ , and its formula has a plus: $\cos A\cos B+\sin A\sin B$ .

4Double angles

$\sin2A=2\sin A\cos A$

$\tan2A=\dfrac{2\tan A}{1-\tan^2A}$

$\cos2A=\cos^2A-\sin^2A$

$\cos2A=1-2\sin^2A=2\cos^2A-1$

Example $\sin A=3/5$ in QII: $\cos A=-4/5$ , $\sin2A=-24/25$ , $\cos2A=7/25$ .

Trap $\sin2A\ne2\sin A$ : for $\sin A=2/3$ , $2\sin A=4/3$ , which no sine can equal.

5Proving identities

$\dfrac{\sin2x}{1+\cos2x}=\dfrac{2\sin x\cos x}{2\cos^2x}=\tan x$

UseWork on one side at a time. Rewrite in $\sin$ and $\cos$ , combine fractions, factor, use a Pythagorean swap.

TrapSquaring both sides or moving terms across assumes what you are proving: that is not a proof.

6Solving with identities

UseReduce to one function and one angle, factor, then solve each factor. Never divide by $\cos x$ : factor it out.

Example $\sin2x=\cos x$ : $\cos x(2\sin x-1)=0$ , so $x=\pi/6,\ \pi/2,\ 5\pi/6,\ 3\pi/2$ .

TrapDividing $\sin2x=\cos x$ by $\cos x$ loses $x=\pi/2$ and $3\pi/2$ .

Which $\cos2A$ ?

Only

\sin

elsewhere

1-2\sin^2A

Only

\cos

elsewhere

2\cos^2A-1

Both, or a difference of squares

\cos^2A-\sin^2A

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Core Skills (K)

K1 Reciprocal, Quotient & Cofunction

K2 Pythagorean Identity

K3 Compound Angles

K4 Double Angles

K5 Equations with Identities

Real World (A)

A1 Mural View

A2 Ladder Check

A3 Indoor Kick

Think It Through (T)

T1 Spot the Flaw

T2 Build It Backwards

Boss Level

Telescoping Cosines

Answer key

Core Skills

Real World

Think It Through

Boss Level

Trigonometric Identities Cheat Sheet

Unit Summary

Build your own

Phone card

1Reciprocal, quotient, cofunction

2Pythagorean identities

3Compound angles

4Double angles

5Proving identities

6Solving with identities

Which cos⁡2A\cos2Acos2A?

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Which $\cos2A$ ?