Trigonometric Identities: Precalculus Math Test Answer Key – PDF Download

Precalculus Trig Identities Test Answer Key: questions 1–5

Question 1: Six functions from a point (3 points)

1. \(r=\sqrt{49+576}=25\), so \(\sin\theta=\frac{24}{25}\), \(\cos\theta=-\frac{7}{25}\)
2. \(\tan\theta=-\frac{24}{7}\), \(\text{sec}\,\theta=-\frac{25}{7}\)
3. \(\frac{576}{625}+\frac{49}{625}=\frac{625}{625}=1\). Shown.

Scoring: 1 point for each part; in (a) and (b) both values must be correct.

Question 2: Values from a Pythagorean identity (2 points)

1. \(\cos^{2}\theta=1-\frac{81}{1681}=\frac{1600}{1681}\); cosine is positive in Quadrant IV, so \(\cos\theta=\frac{40}{41}\)
2. \(\tan\theta=\frac{-9/41}{40/41}=\)\(-\frac{9}{40}\)

Scoring: 1 point for each part; part (a) needs the correct sign.

Question 3: Simplify (2 points)

1. \(\tan^{2}\theta\cos^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{2}\theta=\)\(\sin^{2}\theta\)
2. \(\frac{\sin\theta}{\sin^{2}\theta}=\frac{1}{\sin\theta}=\)\(\text{csc}\,\theta\)

Scoring: 1 point for each correct simplification.

Question 4: Verify an identity (3 points)

Start with the right side and multiply top and bottom by \(1+\sin\theta\): \(\frac{\cos\theta(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}=\frac{\cos\theta(1+\sin\theta)}{1-\sin^{2}\theta}=\frac{\cos\theta(1+\sin\theta)}{\cos^{2}\theta}=\frac{1+\sin\theta}{\cos\theta}\), which is the left side. Verified.

Scoring: 1 point for multiplying by the conjugate, 1 point for using \(1-\sin^{2}\theta=\cos^{2}\theta\), 1 point for cancelling and matching the left side.

Question 5: Exact value of sin 105° (2 points)

\(105^{\circ}=60^{\circ}+45^{\circ}\). \(\sin 105^{\circ}=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\)\(\frac{\sqrt{6}+\sqrt{2}}{4}\) (about 0.966).

Scoring: 1 point for a valid split with the correct sum formula, 1 point for the exact simplified result.

Precalculus Trig Identities Test Answer Key: questions 6–8

Question 6: Difference formulas (3 points)

1. \(\cos A=-\frac{15}{17}\) and \(\sin B=-\frac{4}{5}\)
2. \(\cos(A-B)=\cos A\cos B+\sin A\sin B=-\frac{45}{85}-\frac{32}{85}=\)\(-\frac{77}{85}\)
3. \(\sin(A-B)=\sin A\cos B-\cos A\sin B=\frac{24}{85}-\frac{60}{85}=\)\(-\frac{36}{85}\)

Scoring: 1 point for each part; (a) needs both values with correct signs.

Question 7: Double angle from a tangent (3 points)

1. \(\text{sec}^{2}\theta=1+\frac{16}{9}=\frac{25}{9}\), \(\cos\theta< 0\) in Quadrant III, so \(\cos\theta=-\frac{3}{5}\) and \(\sin\theta=-\frac{4}{5}\)
2. \(\sin 2\theta=2(-\frac{4}{5})(-\frac{3}{5})=\)\(\frac{24}{25}\), \(\cos 2\theta=\frac{9}{25}-\frac{16}{25}=\)\(-\frac{7}{25}\)
3. \(\frac{\sin 2\theta}{\cos 2\theta}=\frac{24/25}{-7/25}=\)\(-\frac{24}{7}\) and \(\frac{2\cdot\frac{4}{3}}{1-\frac{16}{9}}=\frac{8/3}{-7/9}=-\frac{24}{7}\). Both agree.

Scoring: 1 point for each part; (c) needs both methods.

Question 8: Prove a double-angle identity (2 points)

Start with the right side and rewrite the tangent: \(\frac{1-\frac{\sin^{2}\theta}{\cos^{2}\theta}}{1+\frac{\sin^{2}\theta}{\cos^{2}\theta}}\). Multiply top and bottom by \(\cos^{2}\theta\) to get \(\frac{\cos^{2}\theta-\sin^{2}\theta}{\cos^{2}\theta+\sin^{2}\theta}=\frac{\cos^{2}\theta-\sin^{2}\theta}{1}=\cos 2\theta\). Proved.

Scoring: 1 point for rewriting in sine and cosine and clearing the fractions, 1 point for using the Pythagorean identity and the double-angle formula.

Scoring summary

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Precalculus Trigonometric Identities test answer key

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