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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 110 (page 488)

Show that $\cos \frac{蟺}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ and use it to find $\sin \frac{蟺}{16}, \cos \frac{蟺}{16}$ .

Short Answer

Expert verified

The value of $\sin \frac{蟺}{16}$ is $\frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2}$ .

The value of $\cos \frac{蟺}{16}$ is $\frac{\sqrt{2 - \sqrt{2 + \sqrt{2}}}}{2}$ .

Step by step solution

01

Step 1. Given information.

Consider the given question,

$\cos \frac{蟺}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

Using half-angle formulas,

$\cos \frac{胃}{2} = 卤 \sqrt{\frac{1 + \cos 胃}{2}} \sin \frac{胃}{2} = 卤 \sqrt{\frac{1 - \cos 胃}{2}}$

Consider $\cos 胃 = \cos (\frac{蟺}{4})$ . Then,

localid="1646421590598" $胃 = \frac{蟺}{4} \frac{胃}{2} = \frac{蟺}{8}$

02

Step 2. Use the half angle formula.

We know that $\cos (\frac{蟺}{4}) = \frac{\sqrt{2}}{2}$ .

Using half-angle formulas,

$\cos \frac{胃}{2} = 卤 \sqrt{\frac{1 + \cos 胃}{2}} \cos \frac{胃}{2} = 卤 \sqrt{\frac{1 + (\frac{\sqrt{2}}{2})}{2}} \cos \frac{胃}{2} = 卤 \sqrt{\frac{2 + \sqrt{2}}{2}}$

As $\frac{胃}{2}$ lies in the first quadrant,

$\cos (\frac{蟺}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2}$

03

Step 3. Use the trigonometric identities.

Consider $\cos 胃 = \cos (\frac{蟺}{8})$ . Then,

role="math" localid="1646421897677" $胃 = \frac{蟺}{8} \frac{胃}{2} = \frac{蟺}{16}$

As $\frac{胃}{2}$ lies in the first quadrant,

$\cos (\frac{蟺}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2}$

Using half-angle formulas,

$\cos \frac{胃}{2} = 卤 \sqrt{\frac{1 + \cos 胃}{2}} \cos \frac{胃}{2} = 卤 \sqrt{\frac{1 + \cos 胃}{2}} \cos \frac{胃}{2} = 卤 \sqrt{\frac{1 + (\frac{\sqrt{2 + \sqrt{2}}}{2})}{2}} \cos \frac{胃}{2} = 卤 \sqrt{\frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2}}$

04

Step 4. Solve for sinπ16.

Consider the given question,

$\sin \frac{胃}{2} = 卤 \sqrt{\frac{1 - \cos 胃}{2}} \sin \frac{胃}{2} = 卤 \sqrt{\frac{1 - (\frac{\sqrt{2 + \sqrt{2}}}{2})}{2}} \sin \frac{胃}{2} = 卤 \frac{\sqrt{2 - \sqrt{2 + \sqrt{2}}}}{2}$

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Most popular questions from this chapter

Express the product $\cos 3 胃 \cos 5 胃$ as a sum containing only sines or only cosines.

$\tan^{2} (胃) - \sec^{2} (胃)$

Show that $\tan^{- 1} (\frac{1}{v}) = \frac{蟺}{2} - \tan^{- 1} v$ , if $v > 0$

If $\tan 伪 = x + 1$ and role="math" localid="1646425111199" $\tan 尾 = x - 1,$ show that

role="math" localid="1646426027733" $2 c o t (伪 - b) = x^{2}$

Establish each identity.

$\frac{\cos^{2} 胃 - \sin^{2} 胃}{1 - \tan^{2} 胃} = \cos^{2} 胃$

See all solutions

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