Q: Solve for $\theta$ in $0 \leq \theta < 2\pi$ if $\tan \theta + \sqrt{3} = 0$.

$\theta = \dfrac{2\pi}{3}, \dfrac{5\pi}{3}$

Q: Verify the identity: $\tan \theta \cos \theta = \sin \theta$. Which of the following justifies the identity?

Because $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$, so $\tan \theta \cos \theta = \sin \theta$

Q: Solve for $x$ in $0 \leq x < 2\pi$: $2\sin^2 x - 3\sin x + 1 = 0$.

$x = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

Q: Find the general solution to $\sin x \cos x + \sin x = 0$.

$x = n\pi$ or $x = \dfrac{\pi}{2} + 2n\pi$, $n \in \mathbb{Z}$

Q: Which equation models the water level $h(t)$ (in feet) at time $t$ (in hours after midnight) for a tide that oscillates between 7 ft and 15 ft, with low tide at 6 AM and high tide at noon, repeating every 12 hours?

$h(t) = 11 + 4\sin\left(\dfrac{\pi}{6}(t - 9)\right)$

Q: Solve for $\theta$ in $0 \leq \theta < 2\pi$: $\cos \theta = \dfrac{1}{2}$.

$\theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$

Q: Which of the following is the correct verification for the identity $\dfrac{\sec^2 \theta - 1}{\sec^2 \theta} = \sin^2 \theta$?

Because $\sec^2 \theta - 1 = \tan^2 \theta$ and $\sec^2 \theta = 1 + \tan^2 \theta$, so $\dfrac{\tan^2 \theta}{1 + \tan^2 \theta} = \sin^2 \theta$

Q: Given $\sin x = \dfrac{1}{4}$ and $x$ is in quadrant I, what is $\sin(2x)$?

$\dfrac{1}{2}\sin(2x) = 2\sin x \cos x = \dfrac{\sqrt{15}}{8}$

Question 1

Find the exact value of $\sin(105^\circ)$ using the sum formula and the fact that $105^\circ = 60^\circ + 45^\circ$.

Accepted Answer

$\dfrac{\sqrt{6} + \sqrt{2}}{4}$

Question 2

Given $\cos(67.5^\circ)$ and $67.5^\circ = \dfrac{135^\circ}{2}$, find the exact value using the half-angle formula.

Accepted Answer

$\dfrac{\sqrt{2 + \sqrt{2}}}{2}$

Question 3

Solve for $	heta$ in $0 \leq 	heta < 2\pi$ if $	an 	heta + \sqrt{3} = 0$.

Accepted Answer

$	heta = \dfrac{2\pi}{3}, \dfrac{5\pi}{3}$

Question 4

If $\sin x = \dfrac{1}{4}$ and $x$ is in quadrant I, what is $\cos(2x)$?

Accepted Answer

$\dfrac{7}{8}$

Question 5

Verify the identity: $	an 	heta \cos 	heta = \sin 	heta$. Which of the following justifies the identity?

Accepted Answer

Because $	an 	heta = \dfrac{\sin 	heta}{\cos 	heta}$, so $	an 	heta \cos 	heta = \sin 	heta$

Question 6

Solve for $x$ in $0 \leq x < 2\pi$: $2\sin^2 x - 3\sin x + 1 = 0$.

Accepted Answer

$x = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

Question 7

Find the exact value: $\sin^{-1}(1)$.

Accepted Answer

$\dfrac{\pi}{2}$

Question 8

Given $\cos \alpha = -\dfrac{1}{3}$ with $\dfrac{\pi}{2} < \alpha < \pi$, and $	an \beta = \dfrac{2}{3}$ with $\pi < \beta < \dfrac{3\pi}{2}$, find $\cos(\alpha - \beta)$.

Accepted Answer

$-\dfrac{1}{3}$

Question 9

Find the general solution to $\sin x \cos x + \sin x = 0$.

Accepted Answer

$x = n\pi$ or $x = \dfrac{\pi}{2} + 2n\pi$, $n \in \mathbb{Z}$

Question 10

Evaluate $	an(\cos^{-1}(\dfrac{5}{13}))$.

Accepted Answer

$\dfrac{12}{5}$

Question 11

Which equation models the water level $h(t)$ (in feet) at time $t$ (in hours after midnight) for a tide that oscillates between 7 ft and 15 ft, with low tide at 6 AM and high tide at noon, repeating every 12 hours?

Accepted Answer

$h(t) = 11 + 4\sin\left(\dfrac{\pi}{6}(t - 9)\right)$

Question 12

Find the exact value: $\sin^{-1}(\cos(\dfrac{\pi}{6}))$.

Accepted Answer

$\dfrac{\pi}{3}$

Question 13

Solve for $	heta$ in $0 \leq 	heta < 2\pi$: $\cos 	heta = \dfrac{1}{2}$.

Accepted Answer

$	heta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$

Question 14

Which of the following is the correct verification for the identity $\dfrac{\sec^2 	heta - 1}{\sec^2 	heta} = \sin^2 	heta$?

Accepted Answer

Because $\sec^2 	heta - 1 = 	an^2 	heta$ and $\sec^2 	heta = 1 + 	an^2 	heta$, so $\dfrac{	an^2 	heta}{1 + 	an^2 	heta} = \sin^2 	heta$

Question 15

Find the exact value: $\cos^{-1}\left(\dfrac{\sqrt{2}}{2}\right)$.

Accepted Answer

$\dfrac{\pi}{4}$

Question 16

If $\sin x = \dfrac{1}{4}$ and $x$ is in quadrant I, what is $	an(2x)$?

Accepted Answer

$\dfrac{4}{3}$

Question 17

Evaluate $\sin(2	heta) + \sqrt{3}\cos 	heta = 0$ for $0 \leq 	heta < 2\pi$. Which is a solution?

Accepted Answer

$	heta = \dfrac{2\pi}{3}$

Question 18

Find the exact value: $	an^{-1}(-1)$.

Accepted Answer

$-\dfrac{\pi}{4}$

Question 19

Given $\sin x = \dfrac{1}{4}$ and $x$ is in quadrant I, what is $\sin(2x)$?

Accepted Answer

$\dfrac{1}{2}\sin(2x) = 2\sin x \cos x = \dfrac{\sqrt{15}}{8}$

Precalculus Trigonometry Review – Step-by-Step Guidance

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