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Chapter 5: Problem 2

Evaluate \(\sin ^{-1} \frac{1}{2}\).

Short Answer

Expert verified
The short answer is: \(\sin^{-1}\frac{1}{2} = \frac{\pi}{6}\) radians or \(30^{\circ}\).

Step by step solution

01

Recall the half-angle formula

Recall that the sine function has the property that \(\sin(\frac{\pi}{6}) = \sin(30^{\circ}) = \frac{1}{2}\). This formula is derived from the Pythagorean theorem and can be applied to the problem at hand.
02

Apply the inverse sine function

Since \(\sin(\frac{\pi}{6}) = \frac{1}{2}\), we can write this equality using the inverse sine function as follows: \[ \sin^{-1}(\sin(\frac{\pi}{6})) = \sin^{-1}(\frac{1}{2}) \] We have \(\theta = \frac{\pi}{6}\), thus \[ \sin^{-1}(\frac{1}{2}) = \theta \]
03

Simplify the expression

Now, we can substitute the angle \(\theta = \frac{\pi}{6}\) back into the equation to get: \[ \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \] So, \(\sin ^{-1} \frac{1}{2}\) evaluates to \(\frac{\pi}{6}\) radians or \(30^{\circ}\).

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Most popular questions from this chapter

Find exact expressions for the indicated quantities, given that $$ \cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2} $$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 4 and 5 in Section 6.3.] $$ \sin \left(-\frac{3 \pi}{8}\right) $$

Find all numbers \(t\) such that $$ \cos ^{-1} t=\sin ^{-1} t $$.

Explain why $$ \cos ^{-1} t=\sin ^{-1} \sqrt{1-t^{2}} $$ whenever \(0 \leq t \leq 1\)

Evaluate \(\sin \left(\cos ^{-1} \frac{1}{3}\right)\)

Suppose \(t\) is such that \(\tan ^{-1} t=\frac{3 \pi}{7}\). Evaluate the following: (a) \(\tan ^{-1} \frac{1}{t}\) (c) \(\tan ^{-1}\left(-\frac{1}{t}\right)\) (b) \(\tan ^{-1}(-t)\)
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