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

`(\cos^2\theta)/(\cos^2\theta)+(\sin^2\theta)/(\cos^2\theta)=\frac1{\cos^2\theta}\iff1+\tan^2\theta=\sec^2\theta`

Because

`\sec\theta=\frac1{\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(\frac{40}9\right)^2}=\frac{41}9`

`\implies\cos\theta=\frac9{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(\frac9{41}\right)^2-1=-(1519)/(1681)`