The derivative of a trigonometric function.

Question

Let: $$f(x) = \sin x + \frac{\sin3x}3 + \frac{\sin5x}5 + \frac{\sin7x}7$$ Prove that: $$f' \left(\frac{\pi}{9}\right) = \frac{1}{2}$$

Could somebody point me in the right direction with this one?

the right direction: the derivative of sinx is cosx. Now can you take the derivative of the other three terms? Don't forget chain rule here. You may need some trig identities to work around with that given value of x — imranfat, Oct 01 '13 at 18:50

score 3 · Answer 1 · edited Apr 13 '17 at 12:19

3

$$f'(x)=\cos x+\cos3x+\cos5x+\cos7x$$

Observe that the angles of cosine ratios are in arithmetic progression

So, using this, we multiply either sides by $\displaystyle2\sin\frac{(3x-x)}2=2\sin x,$

$$2\sin xf'(x)=2\sin x(\cos x+\cos3x+\cos5x+\cos7x)$$

Using $2\sin A\cos B=\sin(A+B)+\sin(A-B)$ $$2\sin xf'(x)=\sin2x+\sin4x-\sin2x+\sin6x-\sin4x+\sin8x-\sin6x=\sin8x$$

If $9x=\pi, 8x=\pi-x\implies \sin8x=\sin(\pi-x)=\sin x\ne0$

Can you take it from here?

edited Apr 13 '17 at 12:19

Community

answered Oct 01 '13 at 18:51

lab bhattacharjee

That's very neat indeed. I need to copy that... – imranfat Oct 01 '13 at 19:42
To be honest, I can't really figure out a way forward from there. I would have approached this problem in a different way. – user98079 Oct 01 '13 at 20:58
What exactly should I do with the 2sinx before the derivative? – user98079 Oct 01 '13 at 21:04
@user98079, observe that I have multiplied either sides after derivative, please have a look into the edited version – lab bhattacharjee Oct 02 '13 at 04:12
I got $$f'(\frac{pi}{9}) = \frac{1}{sin\frac{pi}{9}}$$ but that does not equal 1/2 – user98079 Oct 02 '13 at 04:31
@user98079, how, You should get $$f'\left(\frac\pi9\right)=\frac{\sin\frac{8\pi}9}{2\sin\frac\pi9}$$ $$=\frac{\sin\left(\pi-\frac{\pi}9\right)}{2\sin\frac\pi9}=\frac{\sin\frac\pi9}{2\sin\frac\pi9}=\frac12$$ as $\sin(\pi-x)=\sin x$ – lab bhattacharjee Oct 02 '13 at 04:59

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