- π /2 ≤ θ ≤ π /2. Find the value of θ in radians. = -sqrt 3/2 Write your answer in simplified, rationalized form. Do not round. θ = ______

Identify angle: Identify the angle where the value of cosine is -sqrt 3/2. In the given interval, the cosine function is negative in the second quadrant. So, we look for an angle in the second quadrant where cos(θ) = -sqrt 3/2. The reference angle for which the cosine value is sqrt 3/2 is π/6. Therefore, θ = π - π/6.Find quadrant: Calculate the value of θ. θ = π - π/6 = (6π/6) - (π/6) = 5π/6.

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$\newline$ $- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ . Find the value of $\theta$ in radians. $\newline$ $\theta = -\sqrt{\frac{3}{2}}$ $\newline$ Write your answer in simplified, rationalized form. Do not round. $\newline$ $\theta =$ ______ $\newline$

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Math Problems

Algebra 2

Inverses of sin, cos, and tan: radians

Full solution

\newline

- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

. Find the value of

\theta

in radians.

\newline

\theta = -\sqrt{\frac{3}{2}}

\newline

Write your answer in simplified, rationalized form. Do not round.

\newline

\theta =

______

\newline

Identify angle: Identify the angle where the value of cosine is $-\sqrt{3}/2$ . In the given interval, the cosine function is negative in the second quadrant. So, we look for an angle in the second quadrant where $\cos(\theta) = -\sqrt{3}/2$ . The reference angle for which the cosine value is $\sqrt{3}/2$ is $\pi/6$ . Therefore, $\theta = \pi - \pi/6$ .
Find quadrant: Calculate the value of $\theta$ . $\theta = \pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6}$ .