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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. 62 (page 496)

In Problems , $47 - 68$ establish each identity.
$1 - \frac{1}{2} \sin (2 胃) = \frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃}$

Short Answer

Expert verified

The identity $1 - \frac{1}{2} \sin (2 胃) = \frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃}$ is established.

Step by step solution

01

Step 1 Given identity is,

$1 - \frac{1}{2} \sin (2 胃) = \frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃}$

The formulae used in this problem are,

$\sin (2 胃) = 2 \sin 胃 \cos 胃 a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)$

02

Step 2 Use the formulae in the required places and prove that the values of the expression in both the sides are equal.

$\frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃} = \frac{(\sin 胃 + \cos 胃) (\sin^{2} 胃 + \cos^{2} 胃 - \sin 胃 \cos 胃)}{\sin 胃 + \cos 胃} = \sin^{2} 胃 + \cos^{2} 胃 - \sin 胃 \cos 胃 \frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃} = 1 - \frac{1}{2} \sin (2 胃)$

Hence, proved.

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Most popular questions from this chapter

Rewrite over a common denominator:

$\frac{\sin (胃) + \cos (胃)}{\cos (胃)} + \frac{\sin (胃) + \cos (胃)}{\sin (胃)}$

$\tan \frac{胃}{2} = \frac{1 - \cos 胃}{_____}$ .

Rewrite over a common denominator:

$\frac{1}{1 - \cos (v)} + \frac{1}{1 + \cos (v)}$

solve each equation on the interval

$0 鈮� 胃 < 2 蟺$

$\sin 胃 + \cos 胃 = \sqrt{2}$

True or False

$\tan (2 胃) = \frac{2 \tan 胃}{1 - \tan^{2} 胃}$

See all solutions

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