# Derivative proof using sine and cosine

• May 1st 2011, 05:09 AM
RezMan
Derivative proof using sine and cosine
Hi guys can you please help me with this problem:

If n is a positive integer, prove this derivative:

(sin^n)xcosnx = n(sin^n-1)xcos(n+1)x

how the heck do I do that?
• May 1st 2011, 05:27 AM
topsquark
Quote:

Originally Posted by RezMan
Hi guys can you please help me with this problem:

If n is a positive integer, prove this derivative:

(sin^n)xcosnx = n(sin^n-1)xcos(n+1)x

how the heck do I do that?

First, is your problem
$sin^{(n)}(x)~cos(nx) = n(sin^{(n)}(x)~cos((n + 1)x)$

where
$sin^{(n)}(x)$

represents the nth derivative of sin(x)? If so, note that the theorem is not true for n = 1.

-Dan
• May 1st 2011, 06:04 AM
HallsofIvy
My interpretation, though it is written badly, is to show that the derivative of f(x)= sin^n(x)cos(nx) is f'(x)= n sin^{n-1}(x)cos((n+1)x).

To start with use the product rule and chain rule:
f'(x)= n sin^{n-1}(x)cos(x)cos(nx)+ sin^n(x)(-n sin(nx))= n sin^{n-1}(x)[cos(x)cos(nx)- sin(x)sin(nx)]

Now use the trig identity cos(a+b)= cos(a)cos(b)- sin(a)sin(b) to rewrite the quantitiy inside the [ ].
• May 1st 2011, 06:11 AM
topsquark
As psychic as ever. (Nod) Good call.

-Dan
• May 1st 2011, 10:32 PM
RezMan
Thank you so much! But I still need some help. Please bear with me. Also, topsquark, can you show me how you made the equation look like that on this board?

I get that (sin^n)x turns into n(sin^n-1) but can you explain in more detail what's going on here cos(x)cos(nx)+ sin^n(x)(-n sin(nx)). Like I don't quite see where the chain rule is applied.
• May 2nd 2011, 03:07 AM
topsquark
Quote:

Originally Posted by RezMan
Also, topsquark, can you show me how you made the equation look like that on this board?

You can visit the Latex Help forum to learn how to use the LaTeX commands. For the moment LaTeX is down but there is a temporary fix that I like. Instead of using the [tex] tags, use [tex].

-Dan
• May 5th 2011, 08:28 PM
RezMan
thanks guys!