# Thread: Derivative proof using sine and cosine

1. ## Derivative proof using sine and cosine

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?

2. Originally Posted by RezMan

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?
$sin^{(n)}(x)~cos(nx) = n(sin^{(n)}(x)~cos((n + 1)x)$

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

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

-Dan

3. 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).

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 [ ].

4. As psychic as ever. Good call.

-Dan

5. 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.

6. 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

7. thanks guys!

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