/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Find an exact expression for \(\... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 65

Find an exact expression for \(\sin \frac{\pi}{24}\).

Short Answer

Expert verified
The exact expression for \(\sin \frac{\pi}{24}\) is \(\frac{\sqrt{6} - \sqrt{2}}{4}\).

Step by step solution

01

Rewrite the angle

Rewrite \(\frac{\pi}{24}\) as the difference of known angles: \[\frac{\pi}{24} = \frac{\pi}{4} - \frac{\pi}{6}\]
02

Use the angle subtraction formula for sine

The angle subtraction formula for sine is: \[\sin (A - B) = \sin A \cos B - \cos A \sin B\] Using this formula, with \(A = \frac{\pi}{4}\) and \(B = \frac{\pi}{6}\), we get: \[\sin \frac{\pi}{24} = \sin \frac{\pi}{4} \cos \frac{\pi}{6} - \cos \frac{\pi}{4} \sin \frac{\pi}{6}\]
03

Find the sine and cosine values for \(\frac{\pi}{4}\) and \(\frac{\pi}{6}\)

Recall the basic values of sine and cosine for these angles: \[\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \qquad \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\] \[\sin \frac{\pi}{6} = \frac{1}{2}, \qquad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\]
04

Substitute the values into the angle subtraction formula

Replace the sine and cosine values in the formula: \[\sin \frac{\pi}{24} = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)\]
05

Simplify the expression

Now, simplify the expression to find the exact value of \(\sin \frac{\pi}{24}\): \[\sin \frac{\pi}{24} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}\] So, the exact expression for \(\sin \frac{\pi}{24}\) is \(\frac{\sqrt{6} - \sqrt{2}}{4}\).

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Most popular questions from this chapter

Show that \(\cos 20^{\circ}\) is a zero of the polynomial \(8 x^{3}-6 x-1\) [Hint: Set \(\theta=20^{\circ}\) in the identity from the previous problem.]

Show that $$\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$$ for all \(x, y\). [Hint: Take \(u=\frac{x+y}{2}\) and \(v=\frac{x-y}{2}\) in the formula given by Example 5 .]

Find a formula for \(\tan \left(\theta-\frac{\pi}{2}\right)\).

Evaluate the indicated expressions assuming that $$ \begin{array}{ll} \cos x=\frac{1}{3} & \text { and } \sin y=\frac{1}{4} \\ \sin u=\frac{2}{3} & \text { and } \cos v=\frac{1}{5} \end{array} $$ Assume also that \(x\) and \(u\) are in the interval \(\left(0, \frac{\pi}{2}\right),\) that \(y\) is in the interval \(\left(\frac{\pi}{2}, \pi\right),\) and that \(v\) is in the interval \(\left(-\frac{\pi}{2}, 0\right) .\) $$\sin (x+y)$$

Show that $$\cos \frac{\pi}{32}=\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}$$ [Hint: First do Exercise 66.]
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