Take this trigonometric function:

$\displaystyle \frac{d \(sin{(\cos{\theta})})^2}{d \theta}}$

Using the chain rule i think it becomes:

$\displaystyle -2\theta\sin{(\cos{\theta})}\sin{\theta}$

But if we use the identity:

$\displaystyle \sin^2{\theta} = 1-\cos^2{\theta}$

then the equation becomes

$\displaystyle \frac{d (1-\theta^2)}{d \theta}} = -2\theta$

But if,

$\displaystyle -2\theta = -2\theta\sin{(\cos{\theta})}\sin{\theta}$

then,

$\displaystyle \sin{(\cos{\theta})}\sin{\theta} = 1 $

Which doesn't appear so. Why am I getting different answers for the same derivative? Did I do something wrong in the above steps?

I accidentally took :

$\displaystyle \cos(\cos\theta) = \theta$

P.S. This is my first post and using [tex] is VERY slow and irritating, is there a faster way to insert math expressions