Can someone show me an induction for $$ \sum_{n}^{M} \cos(2n) = \frac{\sin(M) \cos(M+1)}{\sin(1)} $$? My problem is doing that induction with $M$, I am not sure how to proceed to get the right side of equation.
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If $ \sum_{n}^{M} \cos(2n) = \dfrac{\sin(M) \cdot\cos(M+1)}{\sin(1)} ,$
$ \sum_n^{M+1} \cos(2n) = \dfrac{\sin(M) \cdot\cos(M+1)}{\sin(1)}+\cos2(M+1) $
$$=\frac{2\sin(M) \cdot\cos(M+1)+2\sin1\cdot\cos2(M+1)}{2\sin1}$$
Using Werner Formula $[2\sin B\cos A=\sin(A+B)-\sin(A-B)],$ this becomes
$$\frac{\sin(2M+1)-\sin1+\sin(2M+3)-\sin(2M+1)}{2\sin1}$$
$$=\frac{\sin(2M+3)-\sin1}{2\sin1}$$
Using Prosthaphaeresis Formula $(\sin C-\sin D),$ this becomes
$$=\frac{2\sin(M+1)\cos(M+2)}{2\sin1}$$
lab bhattacharjee
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Oh, I see! My problem was to add that $cos2(M+1)$. Thank you, I will accept as an answer as soon as I can – Superian007 Nov 25 '14 at 09:55
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@Superian007, My pleasure. A little amelioration incorporated. – lab bhattacharjee Nov 25 '14 at 09:56
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@Superian007, See also : http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – lab bhattacharjee Nov 25 '14 at 09:57
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For $n=M+1$, $\dfrac{\sin(M).\cos(M+1)}{\sin1}+\cos(2M+2)=\cos(M+1)[\dfrac{\sin M}{\sin 1}+2\cos(M+1)]-1=\cos(M+1)[\dfrac{\sin M+\sin(M+2)-\sin(M)}{\sin1}]-1=\dfrac{\cos(M+1)\sin(M+2)}{\sin1}-1=\dfrac{\cos(M+1)\sin(M+2)-\sin((M+2)-(M+1))}{\sin 1}=\dfrac{\cos(M+1)\sin(M+2)-\sin(M+2)\cos(M+1)+\sin(M+1)\cos(M+2)}{\sin1}=\dfrac{\sin(M+1)\cos(M+2)}{\sin1}$
Landon Carter
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