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How did they get from line 4 to 5?

$$\sin{(n+1)\theta} \cdot \cos{n\theta} - \cos{(n+1)\theta} \cdot \sin{n\theta} = \sin{((n+1)\theta-n\theta)}$$

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$\sin\,u\cos\,v-\cos\,u\sin\,v=\sin(u-v)$. Take $u=(n+1)\theta$ and $v=n\theta$... –  J. M. Nov 19 '11 at 12:32
    
Oh man ... I need to revisit my trig identities ... –  Jiew Meng Nov 19 '11 at 12:35

2 Answers 2

up vote 1 down vote accepted

For giggles, here's the scenic route:

Start with

$\sin{(n+1)\theta} \cdot \cos{n\theta} - \cos{(n+1)\theta} \cdot \sin{n\theta}$

and then use the addition formulae:

$\begin{align*}\sin(n+1)\theta&=\sin\,n\theta\cos\,\theta+\cos\,n\theta\sin\,\theta\\\cos(n+1)\theta&=\cos\,n\theta\cos\,\theta-\sin\,n\theta\sin\,\theta\end{align*}$

to yield

$(\sin\,n\theta\cos\,\theta+\cos\,n\theta\sin\,\theta)\cos{n\theta} - (\cos\,n\theta\cos\,\theta-\sin\,n\theta\sin\,\theta)\sin{n\theta}$

after which,

$\begin{align*} &\color{red}{(\sin\,n\theta\cos\,n\theta\cos\,\theta+\cos^2 n\theta\sin\,\theta)} - \color{blue}{(\cos\,n\theta\sin\,n\theta\cos\,\theta-\sin^2 n\theta\sin\,\theta)}\\ &\color{red}{\sin\,n\theta\cos\,n\theta\cos\,\theta}-\color{blue}{\cos\,n\theta\sin\,n\theta\cos\,\theta}+\color{red}{\cos^2 n\theta\sin\,\theta}+\color{blue}{\sin^2 n\theta\sin\,\theta}\\ &(\color{red}{\cos^2 n\theta}+\color{blue}{\sin^2 n\theta})\color{purple}{\sin\,\theta}\\ &\color{purple}{\sin\,\theta} \end{align*}$

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As $J.M.$ has said, one should use the standard trigonometric identities about sums and differences of two variables $a$ and $b$ in a sine of cosine function. In fact whenever you see expressions like that it is very useful to know the following formulas, which I always remember as:

$\sin (a \pm b) = \sin a \cos b \pm \cos a \sin b,$

$\cos (a \pm b) = \cos a \cos b \mp \sin a \sin b$.

If you use the first of the two (the one about $\sin (a - b)$) as J.M. suggested by substituting in $a = (n+1) \theta$, $b = n \theta$, your answer pops out immediately. A little bit more about their usefulness:

Sometimes say you have an inner product space of finite dimension, say $W$ that is the span of $\{1, \cos x, \sin x\}$. If I asked you to calculate the projection of the function $\cos(x - \alpha)$, $\alpha \in [0, \pi]$ onto the subspace $W$ with respect to the inner product $\langle, \rangle$ given by

$\langle f(x), g(x) \rangle = \int_{-\pi/2}^{\pi/2} f(x)g(x) dx$,

you would need to evaluate things like say $\int_{-\pi/2}^{\pi/2} \cos x \cos (x - \alpha) dx$. Now the formulas I told you of above will come very handy in evaluating this integral.

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Could the downvoter please explain the reason for the downvote? –  user38268 Nov 19 '11 at 13:31
1  
I think your example is not good for describing importance and application of trigonometry. And mentioning infinite dimensional spaces inner product etc won't give more understanding for jiewmeng, because he is not familiar with this notions (I'm sure). –  Norbert Nov 19 '11 at 13:46
    
@Norbert I accept the reason for your downvote. –  user38268 Nov 19 '11 at 21:37

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