Chapter 7: Q. 78 (page 458)

In Problems 67–78,
$f (x) = \sin x, - \frac{π}{2} \leq x \leq \frac{π}{2}, g (x) = \cos x, 0 \leq x \leq π and h (x) = tanx, - \frac{π}{2} < x < \frac{π}{2}$
Find the exact value of each composite function.
$h (f^{- 1} (- \frac{2}{5}))$

Short Answer

Expert verified

The required exact value of given expression is $\frac{- 2}{\sqrt{21}}$ .

Step by step solution

Step 1. Given information.

The composite function $h (f^{- 1} (- \frac{2}{5}))$ .

Step 2. Here, let f-1-25=x.

So,

$f (x) = - \frac{2}{5} \sin x = - \frac{2}{5} \cos x = \sqrt{1 - {(- \frac{2}{5})}^{2}} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$

Step 3. Now, find hf-1-25.

$h (f^{- 1} (- \frac{2}{5})) = h (x) = \tan x = \frac{\sin x}{\cos x} = \frac{- 2 / 5}{\sqrt{21} / 5} = \frac{- 2}{\sqrt{21}}$

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In Problems 67–78,
$f (x) = \sin x, - \frac{π}{2} \leq x \leq \frac{π}{2}, g (x) = \cos x, 0 \leq x \leq π and h (x) = tanx, - \frac{π}{2} < x < \frac{π}{2}$
Find the exact value of each composite function.
$h (f^{- 1} (- \frac{2}{5}))$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Here, let f-1-25=x.

Step 3. Now, find hf-1-25.

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In Problems 67–78,fx=sinx,-π2≤x≤π2,gx=cosx,0≤x≤πandhx=tanx,-π2<x<π2Find the exact value of each composite function.hf-1-25

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Here, let f-1-25=x.

Step 3. Now, find hf-1-25.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

Applied Mathematics

Pure Maths

Mechanics Maths

Statistics

Study anywhere. Anytime. Across all devices.

In Problems 67–78,
$f (x) = \sin x, - \frac{π}{2} \leq x \leq \frac{π}{2}, g (x) = \cos x, 0 \leq x \leq π and h (x) = tanx, - \frac{π}{2} < x < \frac{π}{2}$
Find the exact value of each composite function.
$h (f^{- 1} (- \frac{2}{5}))$