Sec 6.6 Inverse Trig: Problem 5 (1 point) Evaluate the following expressions. Your answer must be an angle in radians and in the interval $$ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$. (a) $$ \sin ^{-1}(-1)= $$ $$ \square $$ (b) $$ \sin ^{-7}\left(\frac{\sqrt{2}}{2}\right)= $$ $$ \square $$ (c) $$ \sin ^{-1}\left(\frac{x}{2}\right)= $$ $$ \square $$ Note: Wou cun earn partiol credit on this problem.

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Let's evaluate each expression step by step.

### (a) $$ \sin^{-1}(-1) $$

The function $$ \sin^{-1}(x) $$ (also known as arcsin) gives us the angle whose sine is $$ x $$. We need to find the angle $$ \theta $$ such that:

$$
\sin(\theta) = -1
$$

The sine function equals $$-1$$ at:

$$
\theta = -\frac{\pi}{2}
$$

Thus,

$$
\sin^{-1}(-1) = -\frac{\pi}{2}
$$

### (b) $$ \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) $$

Again, we are looking for the angle $$ \theta $$ such that:

$$
\sin(\theta) = \frac{\sqrt{2}}{2}
$$

The sine function equals $$\frac{\sqrt{2}}{2}$$ at:

$$
\theta = \frac{\pi}{4}
$$

Thus,

$$
\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}
$$

### (c) $$ \sin^{-1}\left(\frac{x}{2}\right) $$

For this expression, we are looking for the angle $$ \theta $$ such that:

$$
\sin(\theta) = \frac{x}{2}
$$

The output of $$ \sin^{-1}\left(\frac{x}{2}\right) $$ will depend on the value of $$ x $$. The range of $$ \sin^{-1}(x) $$ is defined for $$ x $$ in the interval $$[-1, 1]$$. Therefore, $$ \frac{x}{2} $$ must also be in the interval $$[-1, 1]$$, which means $$ x $$ must be in the interval $$[-2, 2]$$.

Thus, the answer for this part is:

$$
\sin^{-1}\left(\frac{x}{2}\right) = \theta \quad \text{where } \theta \text{ is such that } \sin(\theta) = \frac{x}{2}
$$

### Summary of Answers

(a) $$ \sin^{-1}(-1) = -\frac{\pi}{2} $$

(b) $$ \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} $$

(c) $$ \sin^{-1}\left(\frac{x}{2}\right) = \theta $$ where $$ \sin(\theta) = \frac{x}{2} $$ and $$ x \in [-2, 2] $$
(a) $$ \sin^{-1}(-1) = -\frac{\pi}{2} $$ (b) $$ \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} $$ (c) $$ \sin^{-1}\left(\frac{x}{2}\right) = \theta $$ where $$ \sin(\theta) = \frac{x}{2} $$ and $$ x \in [-2, 2] $$

Sec 6.6 Inverse Trig: Problem 5
(1 point)

Evaluate the following expressions. Your answer must be an angle in radians and in the interval .
(a)
(b)
©

Note: Wou cun earn partiol credit on this problem.

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(a)

(b)

©

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Sec 6.6 Inverse Trig: Problem 5 (1 point) Evaluate the following expressions. Your answer must be an angle in radians and in the interval . (a) (b) © Note: Wou cun earn partiol credit on this problem.