Trigonometric Identities and Equations: practice solutions, Grade 12 – download the PDF

1 Simplify with Pythagorean identities ★★★

a. From \(\sin^2x+\cos^2x=1\) we get \(1-\cos^2x=\sin^2x\).

b. From \(1+\tan^2x=\sec^2x\) we get \(\sec^2x-1=\tan^2x\).

c. From \(1+\cot^2x=\csc^2x\) we get \(\csc^2x-\cot^2x=1\).

2 Cosine given, sine and tangent wanted ★★★

\(\sin^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}\), so \(\sin\theta=\pm\dfrac{12}{13}\). Sine is negative in Quadrant IV, so \(\sin\theta=-\dfrac{12}{13}\).

Then \(\tan\theta=\dfrac{-12/13}{5/13}=-\dfrac{12}{5}\).

3 Exact value of sin 15 degrees ★★★

\(\sin(45^\circ-30^\circ)=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac12=\dfrac{\sqrt6-\sqrt2}{4}\).

Check: \(\approx0.259\), which is a plausible value for a small angle.

4 Cosine of a double angle ★★★

Use the form that contains only sine: \(\cos2\theta=1-2\sin^2\theta=1-2\cdot\dfrac19=\dfrac79\).

5 True or false: sine of a sum ★★★

False. Take \(A=B=30^\circ\). Then \(\sin(A+B)=\sin60^\circ=\dfrac{\sqrt3}{2}\approx0.866\), but \(\sin A+\sin B=\dfrac12+\dfrac12=1\). One counterexample is enough to disprove it. The correct formula is \(\sin A\cos B+\cos A\sin B\).

6 Exact value of sin 105 degrees ★★★

\(\sin105^\circ=\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}+\dfrac12\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt6+\sqrt2}{4}\approx0.966\).

7 Tangent of a double angle ★★★

\(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac12}{1-\frac14}=\dfrac{1}{3/4}=\dfrac43\).

8 A basic equation on one turn ★★★

The reference angle with sine \(\dfrac{\sqrt3}{2}\) is \(\dfrac{\pi}{3}\). Sine is positive in Quadrants I and II, so \(x=\dfrac{\pi}{3}\) or \(x=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}\).

9 Cosine of 15 degrees from a difference ★★★

\(\cos\left(\dfrac\pi3-\dfrac\pi4\right)=\cos\dfrac\pi3\cos\dfrac\pi4+\sin\dfrac\pi3\sin\dfrac\pi4=\dfrac12\cdot\dfrac{\sqrt2}{2}+\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt2+\sqrt6}{4}\).

That is about 0.966, close to 1, as expected for a small angle.

10 Tangent of 15 degrees by half-angle ★★★

\(\tan15^\circ=\dfrac{1-\cos30^\circ}{\sin30^\circ}=\dfrac{1-\frac{\sqrt3}{2}}{\frac12}=2-\sqrt3\approx0.268\).

11 Half-angle with a given cosine ★★★

Since \(45^\circ<\dfrac\theta2<90^\circ\), both are positive.

\(\sin\dfrac\theta2=\sqrt{\dfrac{1+\frac{7}{25}}{2}}=\sqrt{\dfrac{16}{25}}=\dfrac45\) and \(\cos\dfrac\theta2=\sqrt{\dfrac{1-\frac{7}{25}}{2}}=\sqrt{\dfrac{9}{25}}=\dfrac35\).

Check: \(\cos\theta=\cos^2\dfrac\theta2-\sin^2\dfrac\theta2=\dfrac{9-16}{25}=-\dfrac7{25}\).

12 Product-to-sum practice ★★★

a. \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\), so \(2\sin5x\cos3x=\sin8x+\sin2x\).

b. \(\cos75^\circ\cos15^\circ=\tfrac12\left[\cos90^\circ+\cos60^\circ\right]=\tfrac12\left(0+\dfrac12\right)=\dfrac14\).

13 Sum-to-product practice ★★★

a. \(\sin7x+\sin3x=2\sin\dfrac{7x+3x}{2}\cos\dfrac{7x-3x}{2}=2\sin5x\cos2x\).

b. \(\cos75^\circ+\cos15^\circ=2\cos45^\circ\cos30^\circ=2\cdot\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt6}{2}\approx1.225\).

14 Quadratic in sine ★★★

Factor: \((2\sin x+1)(\sin x-1)=0\), so \(\sin x=-\dfrac12\) or \(\sin x=1\).

\(\sin x=1\) gives \(x=\dfrac\pi2\). \(\sin x=-\dfrac12\) gives angles in Quadrants III and IV: \(x=\dfrac{7\pi}{6}\) and \(x=\dfrac{11\pi}{6}\).

Solutions: \(\dfrac\pi2,\ \dfrac{7\pi}6,\ \dfrac{11\pi}6\).

15 Verify tan + cot ★★★

Work on the left side: \(\tan x+\cot x=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=\dfrac{\sin^2x+\cos^2x}{\sin x\cos x}=\dfrac{1}{\sin x\cos x}=\sec x\csc x\). The sides match.

16 Verify a half-angle quotient ★★★

Use \(\sin2x=2\sin x\cos x\) and \(1+\cos2x=1+(2\cos^2x-1)=2\cos^2x\).

\(\dfrac{2\sin x\cos x}{2\cos^2x}=\dfrac{\sin x}{\cos x}=\tan x\). Verified.

17 Equation with cos 2x and cos x ★★★

Replace \(\cos2x\) by \(2\cos^2x-1\): \(2\cos^2x-\cos x-1=0\), so \((2\cos x+1)(\cos x-1)=0\).

\(\cos x=1\) gives \(x=0\). \(\cos x=-\dfrac12\) gives \(x=\dfrac{2\pi}{3}\) and \(x=\dfrac{4\pi}{3}\).

Solutions: \(0,\ \dfrac{2\pi}3,\ \dfrac{4\pi}3\).

18 Do not divide by cosine ★★★

Write \(2\sin x\cos x-\sqrt3\cos x=0\), then factor: \(\cos x\,(2\sin x-\sqrt3)=0\).

\(\cos x=0\) gives \(x=\dfrac\pi2,\ \dfrac{3\pi}2\). \(\sin x=\dfrac{\sqrt3}{2}\) gives \(x=\dfrac\pi3,\ \dfrac{2\pi}3\).

Solutions: \(\dfrac\pi3,\ \dfrac\pi2,\ \dfrac{2\pi}3,\ \dfrac{3\pi}2\). Dividing by \(\cos x\) would have lost two of them.

19 Launch angle of a soccer ball ★★★

\(30=\dfrac{400\sin2\theta}{9.8}\), so \(\sin2\theta=\dfrac{30\cdot9.8}{400}=0.735\).

With a calculator, \(2\theta\approx47.3^\circ\) or \(2\theta\approx180^\circ-47.3^\circ=132.7^\circ\).

Therefore \(\theta\approx23.7^\circ\) or \(\theta\approx66.3^\circ\): a low, fast kick and a high, slow kick travel the same distance. Their angles add up to \(90^\circ\).

20 Two electrical signals ★★★

Use the sum-to-product formula: \(\sin\left(x+\dfrac\pi3\right)+\sin\left(x-\dfrac\pi3\right)=2\sin x\cos\dfrac\pi3=2\sin x\cdot\dfrac12=\sin x\).

Alternatively expand both with the sum formula: the terms \(\pm\cos x\sin\dfrac\pi3\) cancel and \(2\sin x\cos\dfrac\pi3=\sin x\) remains.

21 Triple-angle formula ★★★

\(\cos3x=\cos2x\cos x-\sin2x\sin x=(2\cos^2x-1)\cos x-2\sin^2x\cos x\).

Replace \(\sin^2x\) by \(1-\cos^2x\): \(=2\cos^3x-\cos x-2\cos x+2\cos^3x=4\cos^3x-3\cos x\). Proved.

22 Extraneous solutions ★★★

Squaring: \(\sin^2x+2\sin x\cos x+\cos^2x=1\), so \(1+\sin2x=1\) and \(\sin2x=0\). Candidates: \(x=0,\ \dfrac\pi2,\ \pi,\ \dfrac{3\pi}2\).

Check in the original equation: \(x=0\): \(0+1=1\) yes. \(x=\dfrac\pi2\): \(1+0=1\) yes. \(x=\pi\): \(0-1=-1\) no. \(x=\dfrac{3\pi}2\): \(-1+0=-1\) no.

Solutions: \(0\) and \(\dfrac\pi2\). Squaring can introduce false solutions, so always check.