## anonymous 5 years ago cos(2tan-1(x))

1. anonymous

its arc-tan, the -1 is suppose to be a ^-1

2. anonymous

must use $cos(2\theta)=cos^2(\theta)-sin^2(\theta)$ so you need $cos(tan^{-1}(x))$ and $sin(tan^{-1}(x))$

3. anonymous

do you know how to find them?

4. anonymous

no

5. anonymous

ok do this { draw a right triangle and label one angle as theta, theta =$tan^{-1}(x)$ the angle whose tangent is x

6. anonymous

since tangent is $\frac{opp}{adj}$ label the opposite angle x and the adjacent angle as 1 so you have $\frac{x}{1}=x$ as the tangent of your angle

7. anonymous

by pythagoras the hypotenuse is$\sqrt{x^2+1}$

8. anonymous

$sin(tan^{-1}(x))=\frac{opp}{hyp}=\frac{x}{\sqrt{x^2+1}}$

9. anonymous

$cos(tan^{-1}(x))=\frac{adj}{hyp}=\frac{1}{\sqrt{x^2+1}}$

10. anonymous

now use the "double angle" formula to find $cos(2tan^{-1}(x))$

11. anonymous

it is the top line i wrote. calling $tan^{-1}(x)=\theta$ it is $cos^2(\theta)-sin^2(\theta)$

12. anonymous

you get $\frac{1}{x^2+1}-\frac{x^2}{x^2+1}=\frac{1-x^2}{x^2+1}$

13. anonymous

the important part is being able to find $sin(tan{-1}(x))$ or $sin(tan{-1}(x))$

14. anonymous

i think i made a mistake on the last post where i said $cos(tan^{-1}(x))$ was something else. it is $\frac{1}{\sqrt{x^2+1}}$