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Chapter 5: Problem 47

Verify the identity. $$\tan \left(\sin ^{-1} x\right)=\frac{x}{\sqrt{1-x^{2}}}$$

Short Answer

Expert verified
The short answer is that the provided identity is correct.

Step by step solution

01

Rewrite using a triangle

Start off by letting \(x = \sin(y)\), for which the inverse \(y = \sin^{-1}(x)\). Now this can be interpreted as an angle in a right triangle, and by definition of sine \( \sin(y) = x = \frac{opposite}{hypotenuse}\). Using Pythagoras theorem, we find \( \cos(y) = \frac{adjacent}{hypotenuse} = \frac{\sqrt{1 - opposite^{2}}}{hypotenuse} = \sqrt{1 - x^{2}}\).
02

Substitute back into the equation

From the definitions above, the expression \( \tan(\sin^{-1}(x)) = \tan(y) = \frac{\sin(y)}{\cos(y)} = \frac{x}{\sqrt{1 - x^{2}}}\).
03

Final check

Check that this final expression is same as given in the original problem, which it is.

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Most popular questions from this chapter

True or False? In Exercises \(81-84\) , determine whether the statement is true or false. Justify your answer. \(\tan \left(x-\frac{\pi}{4}\right)=\frac{\tan x+1}{1-\tan x}\)

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Solving a Multiple-Angle Equation, find the exact solutions of the equation in the interval \(0,2 \pi )\) $$(\sin 2 x+\cos 2 x)^{2}=1$$
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