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

1 Pythagorean identities / 4 pts

a. \(\sec^2\theta=1+\dfrac{64}{225}=\dfrac{289}{225}\); in Quadrant III cosine is negative, so \(\sec\theta=-\dfrac{17}{15}\) and \(\cos\theta=-\dfrac{15}{17}\). (1 pt)

b. \(\sin\theta=\tan\theta\cos\theta=\dfrac{8}{15}\cdot\left(-\dfrac{15}{17}\right)=-\dfrac{8}{17}\). (1 pt)

c. \(\sin2\theta=2\left(-\dfrac8{17}\right)\left(-\dfrac{15}{17}\right)=\dfrac{240}{289}\). (1 pt)

d. \(\cos2\theta=\dfrac{225}{289}-\dfrac{64}{289}=\dfrac{161}{289}\). (1 pt)

2 Exact values / 4 pts

a. \(\sin(45^\circ+30^\circ)=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12=\dfrac{\sqrt6+\sqrt2}{4}\). (2 pts)

b. \(22.5^\circ\) is in Quadrant I, so \(\cos22.5^\circ=\sqrt{\dfrac{1+\cos45^\circ}{2}}=\sqrt{\dfrac{2+\sqrt2}{4}}=\dfrac{\sqrt{2+\sqrt2}}{2}\approx0.924\). (2 pts)

3 Verifying identities / 4 pts

a. Multiply top and bottom of the left side by \(1+\sin x\): \(\dfrac{\cos x(1+\sin x)}{1-\sin^2x}=\dfrac{\cos x(1+\sin x)}{\cos^2x}=\dfrac{1+\sin x}{\cos x}\). (2 pts)

b. \(\cos2x=\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x)\); dividing by \(\cos x+\sin x\) leaves \(\cos x-\sin x\). (2 pts)

4 A trigonometric equation / 4 pts

Replace \(\sin^2x\) by \(1-\cos^2x\): \(2-2\cos^2x+3\cos x=0\), that is \(2\cos^2x-3\cos x-2=0\). (1 pt)

Factor: \((2\cos x+1)(\cos x-2)=0\). (1 pt)

\(\cos x=2\) is impossible because cosine never exceeds 1. (1 pt)

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

5 Products and sums / 4 pts

a. \(\sin3x\cos x=\tfrac12\left[\sin4x+\sin2x\right]\). (2 pts)

b. \(\sin5x-\sin x=2\cos3x\sin2x=0\). (1 pt)

\(\cos3x=0\) gives \(x=\dfrac\pi6,\ \dfrac\pi2,\ \dfrac{5\pi}6\); \(\sin2x=0\) gives \(x=0,\ \dfrac\pi2,\ \pi\). Solutions: \(0,\ \dfrac\pi6,\ \dfrac\pi2,\ \dfrac{5\pi}6,\ \pi\). (1 pt)