Select the answer which is equivalent to the given expression using your calculator. $\newline$ $\csc \left(\arccos \frac{\sqrt{15}}{4}\right)$ $\newline$ $\frac{4}{\sqrt{15}}$ $\newline$ $4$ $\newline$ $\frac{\sqrt{15}}{4}$ $\newline$ $\sqrt{15}$

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Algebra 2

Simplify expressions using trigonometric identities

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Q. Select the answer which is equivalent to the given expression using your calculator.

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\csc \left(\arccos \frac{\sqrt{15}}{4}\right)

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\frac{4}{\sqrt{15}}

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4

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\frac{\sqrt{15}}{4}

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\sqrt{15}

Understand relationship: Understand the relationship between $\text{arccos}$ and cosine. The $\text{arccos}$ function is the inverse of the cosine function. This means that if $x = \text{arccos}(y)$ , then $y = \cos(x)$ . Therefore, if we have $\text{arccos}(\frac{\sqrt{15}}{4})$ , we can say that $\cos(\theta) = \frac{\sqrt{15}}{4}$ , where $\theta$ is the angle whose cosine is $\frac{\sqrt{15}}{4}$ .
Use Pythagorean identity: Use the Pythagorean identity to find $\sin(\theta)$ . Since $\cos^2(\theta) + \sin^2(\theta) = 1$ , we can find $\sin(\theta)$ by rearranging the identity to $\sin^2(\theta) = 1 - \cos^2(\theta)$ . We know $\cos(\theta) = (\sqrt{15})/4$ , so we can calculate $\sin(\theta)$ as follows: $\newline$ $\sin^2(\theta) = 1 - ((\sqrt{15})/4)^2$ $\newline$ $\sin^2(\theta) = 1 - (15/16)$ $\newline$ $\sin^2(\theta) = (16/16) - (15/16)$ $\newline$ $\sin^2(\theta) = 1/16$ $\newline$ $\cos^2(\theta) + \sin^2(\theta) = 1$ $0$ $\newline$ $\cos^2(\theta) + \sin^2(\theta) = 1$ $1$ or $\cos^2(\theta) + \sin^2(\theta) = 1$ $2$ $\newline$ Since $\cos^2(\theta) + \sin^2(\theta) = 1$ $3$ always gives an angle in the range $\cos^2(\theta) + \sin^2(\theta) = 1$ $4$ , where sine is non-negative, we take the positive value $\cos^2(\theta) + \sin^2(\theta) = 1$ $1$ .
Calculate $\sin(\theta)$ : Calculate $\csc(\theta)$ using the definition of cosecant. Cosecant is the reciprocal of sine, so $\csc(\theta) = \frac{1}{\sin(\theta)}$ . We have found that $\sin(\theta) = \frac{1}{4}$ , so: $\csc(\theta) = \frac{1}{(\frac{1}{4})}$ $\csc(\theta) = 4$
Conclude $\csc(\theta)$ : Conclude that $\csc(\arccos(\frac{\sqrt{15}}{4}))$ is equivalent to $4$ . Since we have calculated $\csc(\theta)$ to be $4$ , where $\theta$ is the angle whose cosine is $\frac{\sqrt{15}}{4}$ , we can say that $\csc(\arccos(\frac{\sqrt{15}}{4})) = 4$ .