# Thread: Various Trig problems involving trigonometric identities

1. ## Various Trig problems involving trigonometric identities

Couldn't think of a title, as i'll be using this thread more than once today for related, but different problems.

for starters:
$\displaystyle \frac{1}{2}\sin{2x}[\tan^2{x}-\cos{2x}+2\cos^2{2x}]$

For this, I ended up with -2sinxcosx, it's supposedly wrong.

Here's my work, for reference:

$\displaystyle \frac{1}{2}\sin{2x}[\tan^2{x}-\cos{2x}+2\cos^2{2x}]$

$\displaystyle \frac{1}{2}\sin{2x}[\tan^2{x}-2\cos^2{x}-1+2\cos^2{2x}]$

$\displaystyle \frac{1}{2}\sin{2x}[\tan^2{x}-1]$

$\displaystyle \sin{x}\cos{x}(\tan^2{x}-1)$

$\displaystyle \sin{x}\cos{x}(\frac{1-\cos{2x}}{1+\cos{2x}}-1)$

$\displaystyle \frac{\sin{x}\cos{x}-\sin{x}\cos{x}\cos{2x}}{1+\cos{2x}}-\sin{x}\cos{x}$

$\displaystyle \frac{-\sin{x}\cos{x}(1+\cos{2x})}{1+\cos{2x}}$

$\displaystyle -2\sin{x}\cos{x}$

2. Only one question:

Originally Posted by zll
...

$\displaystyle \frac{-\sin{x}\cos{x}(1+\cos{2x})}{1+\cos{2x}}$

$\displaystyle -2\sin{x}\cos{x}$ <-- where does the factor 2 comes from?