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Prove the following identities and give the values of \(x\) for which they are true. $$\sin \left(2 \sin ^{-1} x\right)=2 x \sqrt{1-x^{2}}$$

Short Answer

Expert verified
Answer: The identity holds true for all \(x\in[-1,1]\), as that is the domain of the arcsin function.

Step by step solution

01

Apply the double angle formula

The formula for the sine of twice an angle is given by: $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ Let's apply this formula to our given identity, replacing \(\theta\) with \(\sin^{-1}(x)\): $$\sin \left(2 \sin ^{-1} x\right) = 2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x))$$
02

Use the properties of inverse functions

The properties of inverse functions tell us that \(\sin(\sin^{-1}(x)) = x\) as long as x is in the domain of \(\sin^{-1}(x)\). The domain of \(\sin^{-1}(x)\) is \([-1,1]\), so as long as \(x\in[-1,1]\), we can make this substitution: $$2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x)) = 2x\cos(\sin^{-1}(x))$$
03

Find \(\cos(\sin^{-1}(x))\) based on the Pythagorean identity

Recall the Pythagorean identity for sine and cosine: $$\sin^2(\theta) + \cos^2(\theta) = 1$$ With our angle \(\theta = \sin^{-1}(x)\), we have: $$\sin^2(\sin^{-1}(x)) + \cos^2(\sin^{-1}(x)) = 1$$ Substituting \(\sin(\sin^{-1}(x)) = x\), we get: $$x^2 + \cos^2(\sin^{-1}(x)) = 1$$ Now, we can solve for \(\cos(\sin^{-1}(x))\): $$\cos^2(\sin^{-1}(x)) = 1 - x^2$$ $$\cos(\sin^{-1}(x)) = \sqrt{1-x^2}$$
04

Substitute the expression of \(\cos(\sin^{-1}(x))\) into our equation

Now we can substitute the expression for \(\cos(\sin^{-1}(x))\) back into the equation we derived earlier: $$2x\cos(\sin^{-1}(x)) = 2x\sqrt{1-x^2}$$
05

Conclusion and values of x

We've shown that the given identity is true: $$\sin \left(2 \sin ^{-1} x\right)=2 x \sqrt{1-x^{2}}$$ This identity holds true for all \(x\in[-1,1]\), as that is the domain of the arcsin function.

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