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Let $n$ be a odd integer. If $$\sin n\theta=\sum_{r=0}^{n} b_r \sin^r\theta$$ for every value of $\theta$. I have to find $b_1$.

I don't know how to start this?

Thanks.

2 Answers 2

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It's $n$. First, consider this complex identity:

$$e^{in\theta}=(\cos\theta+i\sin\theta)^n$$

$$e^{in\theta}=\cos(n\theta)+i\sin(n\theta)$$

By the binomial development:

$$(\cos\theta+i\sin\theta)^n=\cos^n\theta+in\cos^{n-1}\theta\sin\theta+(powers\,of\,\sin\theta>1)=$$

Because $n$ odd and $n-1$ even:

$$\cos^n\theta+in(\cos^2\theta)^{(n-1)/2}\sin\theta+(powers\,of\,\sin\theta>1)=$$

$$\cos^n\theta+in(1-\sin^2\theta)^{(n-1)/2}\sin\theta+(powers\,of\,\sin\theta>1)=$$

$$\cos^n\theta+in\sin\theta+(powers\,of\,\sin\theta>1)$$

Now:

$$\cos(n\theta)+i\sin(n\theta)=\cos^n\theta+in\sin\theta+(powers\,of\,\sin\theta>1)$$

Taking the imaginary part:

$$\sin(n\theta)=n\sin\theta+ (powers\,of\,\sin\theta>1)$$

And it's done with the bonus of $b_0=0$

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    where did $(1-sin^2 \theta)$ vanish from second step towards third? – 2017-02-13
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    Again using the power, now $(n-1)/2$, of a binomy, the second term is $-((n-1)/2)\sin^{2}\theta$. The next ones have even greater powers of $\sin\theta$. Then, after multiplying by $\sin\theta$, all of them but the first are assimilated to the "tail" of powers of $\sin\theta$ greater than one. – 2017-02-13
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    You are welcome. – 2017-02-13
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$$\sin n\theta=\sum_{r=0}^{n} b_r \sin^r\theta=b_0+b_1\sin \theta+\sum_{r=2}^{n} b_r \sin^r\theta$$ by differentiation of this equivalency we have $$n\cos n\theta=b_1\cos \theta+\sum_{r=2}^{n} b_r r\sin^{r-1}\theta\cos\theta$$ set $\theta=2\pi$ you take $$\color{blue}{b_1=n}$$