• Mar 1st 2013, 06:24 AM
dhtikna
Take this trigonometric function:

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

Using the chain rule i think it becomes:

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

But if we use the identity:

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

then the equation becomes

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

But if,

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

then,

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

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 :

$\cos(\cos\theta) = \theta$(Headbang)

P.S. This is my first post and using [tex] is VERY slow and irritating, is there a faster way to insert math expressions :confused:
• Mar 1st 2013, 06:46 AM
Plato
Quote:

Originally Posted by dhtikna
Take this trigonometric function:
$\frac{d \(sin{(\cos{\theta})})^2}{d \theta}}$

Using the chain rule i think it becomes:
$-2\theta\sin{(\cos{\theta})}\sin{\theta}$

This part of the post is correct. But none of the rest makes any sense.

It is the case that $\sin^2(\cos(\theta))=1-\cos^2(\cos(\theta))$. But that cannot be simplified.

They both have the same derivative.
• Mar 1st 2013, 06:58 AM
HallsofIvy
$cos(arccos(x))= x$ but you have $cos(cos(\theta))$ which is NOT $\theta$.
• Mar 1st 2013, 07:38 AM
dhtikna
I Can't belive i was so stupid. I hadn't been doing trig for a long time so i messed up when I took:

$\cos({\cos\theta}) = \theta$

By the way, any have a answer to my P.S.:

P.S. This is my first post and using [tex] is VERY slow and irritating, is there a faster way to insert math expressions
• Mar 1st 2013, 04:07 PM
Prove It
$\cos({\cos\theta}) = \theta$