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Question Find the value of sin⁡(2tan⁡−1x)\sin \left(2 \tan ^{-1} x\right).

Studdy Solution

STEP 1

Assumptions1. The problem is to simplify the expression sin⁡(tan⁡−1x)\sin \left( \tan ^{-1} x\right). We can use the identity tan⁡−1x=π4−tan⁡−1(1−x1+x)\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left(\frac{1-x}{1+x}\right)3. We can use the double angle formula for sine sin⁡(A)=sin⁡(A)cos⁡(A)\sin(A) =\sin(A)\cos(A)

STEP 2

First, we apply the identity tan⁡−1x=π4−tan⁡−1(1−x1+x)\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left(\frac{1-x}{1+x}\right) to the expression.
sin⁡(2tan⁡−1x)=sin⁡(2(π4−tan⁡−1(1−x1+x)))\sin \left(2 \tan ^{-1} x\right) = \sin \left(2\left(\frac{\pi}{4} - \tan ^{-1} \left(\frac{1-x}{1+x}\right)\right)\right)

STEP 3

implify the expression inside the sine function.
sin⁡(2tan⁡−1x)=sin⁡(π2−2tan⁡−1(1−x1+x))\sin \left(2 \tan ^{-1} x\right) = \sin \left(\frac{\pi}{2} -2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right)

STEP 4

Use the sine of difference identity sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B).
sin⁡(π2−2tan⁡−1(1−x1+x))=sin⁡(π2)cos⁡(2tan⁡−1(1−x1+x))−cos⁡(π2)sin⁡(2tan⁡−1(1−x1+x))\sin \left(\frac{\pi}{2} -2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right) = \sin \left(\frac{\pi}{2}\right)\cos \left(2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right) - \cos \left(\frac{\pi}{2}\right)\sin \left(2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right)

STEP 5

implify the expression using the fact that sin⁡(π2)=1\sin \left(\frac{\pi}{2}\right) =1 and cos⁡(π2)=0\cos \left(\frac{\pi}{2}\right) =0.
sin⁡(π2−2tan⁡−1(1−x1+x))=cos⁡(2tan⁡−1(1−x1+x))\sin \left(\frac{\pi}{2} -2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right) = \cos \left(2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right)

STEP 6

Use the double angle formula for cosine cos⁡(2A)=1−2sin⁡2(A)\cos(2A) =1 -2\sin^2(A).
cos⁡(2tan⁡−1(1−x1+x))=1−2sin⁡2(tan⁡−1(1−x1+x))\cos \left(2\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right) =1 -2\sin^2 \left(\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right)

STEP 7

Use the identity sin⁡(tan⁡−1(A))=A1+A2\sin(\tan^{-1}(A)) = \frac{A}{\sqrt{1 + A^2}}.
1−2sin⁡2(tan⁡−1(1−x1+x))=1−2(1−x1+x1+(1−x1+x)2)21 -2\sin^2 \left(\tan ^{-1} \left(\frac{1-x}{1+x}\right)\right) =1 -2\left(\frac{\frac{1-x}{1+x}}{1 + \left(\frac{1-x}{1+x}\right)^2}\right)^2

STEP 8

implify the expression.
1−2(1−x1+x1+(1−x1+x)2)2=1−x21+x21 -2\left(\frac{\frac{1-x}{1+x}}{1 + \left(\frac{1-x}{1+x}\right)^2}\right)^2 = \frac{1 - x^2}{1 + x^2}So, sin⁡(2tan⁡−1x)=1−x21+x2\sin \left(2 \tan ^{-1} x\right) = \frac{1 - x^2}{1 + x^2}.

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