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Chapter 5: Problem 22

Evaluate \(\cos ^{-1}\left(\sin \frac{4 \pi}{9}\right)\).

Short Answer

Expert verified
The short answer is: \(\cos^{-1}\left(\sin \frac{4\pi}{9}\right) = \frac{\pi}{18}\).

Step by step solution

01

Convert sine to cosine

To convert the sine function into a cosine function, we use the identity \(\cos\left(\frac{\pi}{2} - x\right) = \sin x\). Thus, we can rewrite the given expression as: \[\cos^{-1}\left(\sin\frac{4\pi}{9}\right)=\cos^{-1}\left(\cos\left(\frac{\pi}{2}-\frac{4\pi}{9}\right)\right)\]
02

Find the value of the angle

Now, we need to find the value of \(\frac{\pi}{2}-\frac{4\pi}{9}\): \[\frac{\pi}{2}-\frac{4\pi}{9}= \frac{9\pi}{18}-\frac{8\pi}{18} = \frac{\pi}{18}\]
03

Apply the inverse cosine function

From step 2, we found that the angle is \(\frac{\pi}{18}\). With this knowledge, we can now plug it back into the inverse cosine function from step 1: \[\cos^{-1}\left(\cos\frac{\pi}{18}\right) = \frac{\pi}{18}\] So, the final answer is: \[\cos^{-1}\left(\sin \frac{4\pi}{9}\right) = \frac{\pi}{18}\]

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Most popular questions from this chapter

In 1768 the Swiss mathematician Johann Lambert proved that if \(\theta\) is a rational number in the interval \(\left(0, \frac{\pi}{2}\right),\) then \(\tan \theta\) is irrational. Explain why this result implies that \(\pi\) is irrational.

Suppose \(\theta\) is not an odd multiple of \(\frac{\pi}{2}\). Explain why the point \((\tan \theta, 1)\) is on the line containing the point \((\sin \theta, \cos \theta)\) and the origin.

Explain why $$ \cos ^{-1} t=\tan ^{-1} \frac{\sqrt{1-t^{2}}}{t} $$ whenever \(0

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.] $$ \sin \left(-\frac{5 \pi}{12}\right) $$

Explain why $$ \cos ^{-1} \frac{5}{13}=\sin ^{-1} \frac{12}{13}=\tan ^{-1} \frac{12}{5} $$.
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