# Math Help - Quick question about differentiation of sin

1. ## Quick question about differentiation of sin

How come dy/dx $\sin^2(x)$ is $\sin(2x)$? Thanks.

2. ## Re: Quick question about differentiation of sin

What do you get when you differentiate?

3. ## Re: Quick question about differentiation of sin

Well I did like power rule and chain rule uhh...I'm kinda lost.. I got 2*cos(x) I believe..

4. ## Re: Quick question about differentiation of sin

Yes the power and chain rules are appropriate here, but you lost sin(x) in your result...

$\frac{d}{dx}\left(\sin^2(x) \right)=2\sin^{2-1}(x)\cdot\frac{d}{dx}(\sin(x))=2\sin(x)\cos(x)$

What is the double-angle identity for sine?

5. ## Re: Quick question about differentiation of sin

Originally Posted by Paze
How come dy/dx $\sin^2(x)$ is $\sin(2x)$? Thanks.
\displaystyle \begin{align*} y &= \sin^2{(x)} \\ &= \frac{1}{2} - \frac{1}{2}\cos{(2x)} \\ \\ \frac{dy}{dx} &= 0 - 2\cdot \frac{1}{2} \left[ -\sin{(2x)} \right] \\ &= \sin{(2x)} \end{align*}

6. ## Re: Quick question about differentiation of sin

Thanks Mark. I'm not too good at identifying double angle identities. I was unsure of my method due to the fact that I have a function of x squared. So I'm thinking: ''Is that the power rule or the chain rule?'' Never occured to me that when a function is squared, the function is chain rule and the 'squared' part is power rule. It was intuitive to me, but I didn't quite SEE it. Thanks for clarifying.

Prove It: although your guidance is usually of SUPERB quality, your description confuses me even more this time. You must be applying some identities in your calculations that I am not aware of (or I could be obtuse, sorry if I am).

7. ## Re: Quick question about differentiation of sin

Originally Posted by Paze
Thanks Mark. I'm not too good at identifying double angle identities. I was unsure of my method due to the fact that I have a function of x squared. So I'm thinking: ''Is that the power rule or the chain rule?'' Never occured to me that when a function is squared, the function is chain rule and the 'squared' part is power rule. It was intuitive to me, but I didn't quite SEE it. Thanks for clarifying.

Prove It: although your guidance is usually of SUPERB quality, your description confuses me even more this time. You must be applying some identities in your calculations that I am not aware of (or I could be obtuse, sorry if I am).
My guidance is as superb as it always is, and you are so obtuse you are almost reflex :P but I accept your apology.

Surely you know the double angle identity for cosine...

\displaystyle \begin{align*} \cos{(2x)} &= \cos^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - \sin^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - 2\sin^2{(x)} \\ 2\sin^2{(x)} &= 1 - \cos{(2x)} \\ \sin^2{(x)} &= \frac{1}{2} - \frac{1}{2}\cos{(2x)} \end{align*}

Btw no offence was intended, it's just some light-hearted ribbing

8. ## Re: Quick question about differentiation of sin

Originally Posted by Prove It
My guidance is as superb as it always is, and you are so obtuse you are almost reflex :P but I accept your apology.

Surely you know the double angle identity for cosine...

\displaystyle \begin{align*} \cos{(2x)} &= \cos^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - \sin^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - 2\sin^2{(x)} \\ 2\sin^2{(x)} &= 1 - \cos{(2x)} \\ \sin^2{(x)} &= \frac{1}{2} - \frac{1}{2}\cos{(2x)} \end{align*}

Btw no offence was intended, it's just some light-hearted ribbing
lol. Thanks for clarifying. I see your point now. The thing is, I just completed trigonometry in school now (final test today. It went very well ) so it's not exactly second nature to me. These identities still have to seep in with time. I think I'm better at calculus than trigonometry, since 90% of my math knowledge is self-taught. But that, of course, means that I have a lot of gaps in my mathematical background!

9. ## Re: Quick question about differentiation of sin

As you progress in your studies, you will get many opportunities to fortify your knowledge of pre-calculus concepts.