Precalculus Trigonometric Identities Test Solutions and Scoring (PDF)

Precalculus math test solutions: Trigonometric Identities – Maths PDF

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This trig identities test answers for Precalculus covers trigonometric identities (Common Core F-TF.C.8, F-TF.C.9), with a free PDF to print.

Worked solutions and scoring notes for the 9-question trig identities test. Use the scoring line under each solution to award partial credit.

Precalculus · Common Core State Standards: F-TF.C.8, F-TF.C.9

Trig identities test answers: worked solutions

Question 1 – Simplify [2 pts]

\(\sin^{2}x=1-\cos^{2}x=(1-\cos x)(1+\cos x)\). Cancel \(1-\cos x\).

\(1+\cos x\)

Scoring: 1 point for factoring, 1 point for the result.

Question 2 – Exact value [2 pts]

\(\cos(45^{\circ}+30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\).

\(\frac{\sqrt6-\sqrt2}{4}\)

Scoring: 1 point for the formula with correct values, 1 point for the result.

Question 3 – Double angle [2 pts]

\(\sin\theta=-\frac{12}{13}\). \(\sin2\theta=2\left(-\frac{12}{13}\right)\frac{5}{13}=-\frac{120}{169}\). \(\cos2\theta=\frac{25}{169}-\frac{144}{169}=-\frac{119}{169}\).

\(-\frac{120}{169}\) and \(-\frac{119}{169}\)

Scoring: 1 point each.

Question 4 – Verify [2 pts]

\(1-\cos2x=1-(1-2\sin^{2}x)=2\sin^{2}x\). Then \(\frac{2\sin^{2}x}{2\sin x}=\sin x\).

Verified

Scoring: 1 point for the double-angle substitution, 1 point for simplifying.

Question 5 – Angle addition [3 pts]

Triangles 8-15-17 and 3-4-5.

\(\cos\alpha=\frac{15}{17}\), \(\sin\beta=\frac45\). (a) \(\frac{8}{17}\cdot\frac35+\frac{15}{17}\cdot\frac45=\frac{84}{85}\). (b) \(\frac{15}{17}\cdot\frac35-\frac{8}{17}\cdot\frac45=\frac{13}{85}\). (c) \(\frac{84/85}{13/85}=\frac{84}{13}\).

\(\frac{84}{85},\ \frac{13}{85},\ \frac{84}{13}\)

Scoring: 1 point for each part.

Question 6 – Solve [2 pts]

\(2\sin x\cos x-\cos x=0\), so \(\cos x(2\sin x-1)=0\). \(\cos x=0\): \(\frac{\pi}{2},\frac{3\pi}{2}\). \(\sin x=\frac12\): \(\frac{\pi}{6},\frac{5\pi}{6}\).

\(x=\frac{\pi}{6},\ \frac{\pi}{2},\ \frac{5\pi}{6},\ \frac{3\pi}{2}\)

Scoring: 1 point for factoring, 1 point for all four solutions.

Question 7 – Solve with cos 2x [2 pts]

\(2x=\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3},\frac{11\pi}{3}\), so divide by 2.

\(x=\frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{7\pi}{6},\ \frac{11\pi}{6}\)

Scoring: 1 point for the values of 2x, 1 point for the final solutions.

Question 8 – Projectile range [3 pts]

Same range at 15 and 75 degrees.

(a) \(288\sin30^{\circ}=144\). (b) \(288\sin90^{\circ}=288\). (c) \(\sin2\theta=\sin30^{\circ}\) also holds for \(2\theta=150^{\circ}\), so \(\theta=75^{\circ}\).

\(144\) ft; \(288\) ft; \(75^{\circ}\)

Scoring: 1 point for each part.

Question 9 – Verify a Pythagorean form [2 pts]

\(1+\tan^{2}x=\frac{1}{\cos^{2}x}\), so the product is \(\frac{\cos^{2}x}{\cos^{2}x}=1\).

Verified

Scoring: 1 point for the identity, 1 point for simplifying.

Total score

Total

20 points

Key facts behind this trig identities test answers

Key facts

sin^2 x + cos^2 x = 1 and 1 + tan^2 x = 1/cos^2 x.
tan x = sin x / cos x; csc, sec, cot are reciprocals of sin, cos, tan.
sin(A + B) = sin A cos B + cos A sin B; cos(A + B) = cos A cos B – sin A sin B.
sin 2A = 2 sin A cos A; cos 2A = cos^2 A – sin^2 A = 1 – 2 sin^2 A = 2 cos^2 A – 1.
To verify an identity, work on one side only until it matches the other.

How to use this trig identities test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

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