theta = tan^-1 xθ=tan−1x if and only if -pi/2 < theta < pi/2−π2<θ<π2 and tan theta = xtanθ=x
cos(2theta) = 2cos^2 theta -1cos(2θ)=2cos2θ−1
There are several ways to find cos^2 thetacos2θ for tan theta = xtanθ=x.
Here are two:
Method 1: Sketch a triangle
You can sketch a right triangle with one angle thetaθ. Label the side opposite thetaθ as length xx and the side adjacent has length 11, so the hypotenuse has length sqrt(1+x^2)√1+x2
We can see that cos theta = 1/sqrt(1+x^2)cosθ=1√1+x2
Method 2: Use a trigonometric identity
Recall that tan^2 theta +1 = sec^2 theta = 1/cos^2 thetatan2θ+1=sec2θ=1cos2θ.
So x^2+1 = 1/cos^2 thetax2+1=1cos2θ.
Using either method we continue
cos^2 theta = 1/(1+x^2)cos2θ=11+x2, so
cos(2theta) = 2/(1+x^2) - 1cos(2θ)=21+x2−1
= (1-x^2)/(1+x^2)=1−x21+x2.
And we have
cos(2tan^-1 x) = (1-x^2)/(1+x^2)cos(2tan−1x)=1−x21+x2
