Question

Double angle formulas. Please help?

Answer 1

If $\theta$ is in the first quadrant, we know that $\cos\theta$ is positive.

Recall the Pythagorean identity,

$\cos^2\theta+\sin^2\theta=1$

Dividing both sides by $\cos^2\theta$ gives

$\dfrac{\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}=\dfrac1{\cos^2\theta}\iff1+\tan^2\theta=\sec^2\theta$

Because

$\sec\theta=\dfrac1{\cos\theta}$

if $\cos\theta>0$, then we also have $\sec\theta>0$. This means that we take the square root of both sides in the $\tan$-$\sec$ identity, we take the positive square root and we get

$\sec\theta=\sqrt{1+\left(\dfrac{40}9\right)^2}=\dfrac{41}9$

$\implies\cos\theta=\dfrac9{41}$

Now, recall the double angle identity,

$\cos2\theta=\cos^2\theta-\sin^2\theta$

Apply the Pythagorean identity to eliminate $\sin\theta$:

$\cos2\theta=\cos^2\theta-(1-\cos^2\theta)=2\cos^2\theta-1$

So we end up with

$\cos2\theta=2\left(\dfrac9{41}\right)^2-1=-\dfrac{1519}{1681}$