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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. 102 (page 487)

Show that $\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$

Short Answer

Expert verified

The equation $\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$ can be obtained as

$\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = \cos (\frac{蟺}{2}) \cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$

Step by step solution

01

Step 1. Given data

The given equation for verification is

$\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$

02

Step 2. Proof

As known $\sin^{- 1} (v) + \cos^{- 1} (v) = \frac{蟺}{2} 鈰� (i)$

Use equation i in the given equation

$\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = \cos (\frac{蟺}{2})$

we know $\cos (\frac{蟺}{2}) = 0$

So $\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = \cos (\frac{蟺}{2}) \cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$

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if $\sin 胃 = \frac{4}{5}$ and $胃$ is in quadrant 2, then $\cos 胃 =$ ______

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