Studdy AI Logo
Math
Arithmetic
Fractions and Decimals
Measurement
Geometry
Word Problems
Algebra
Calculus
Statistics
Probability
Number Theory
Discrete Mathematics
Linear Algebra
Differential Equations
Trigonometry
Finance
Solved on Nov 10, 2023

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

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}.

Was this helpful?
banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ContactInfluencer programPolicyTerms
TwitterInstagramFacebookTikTokDiscord