find-the-sum-of-z-sinx-sin2x-sin3x-sinnx-

Question Number 171455 by mokys last updated on 15/Jun/22

f i n d t h e s u m o f z = s i n x + s i n 2 x + s i n 3 x + \dots \dots + s i n n x

Commented by mr W last updated on 15/Jun/22

z = \frac{\cos \frac{x}{2} - \cos (n + \frac{1}{2}) x}{2 \sin \frac{x}{2}}

Answered by mr W last updated on 16/Jun/22

l e t A = \underset{k = 1}{\sum^{n}} \cos k x

l e t B = \underset{k = 1}{\sum^{n}} \sin k x

A + i B = \underset{k = 1}{\sum^{n}} (\cos k x + i \sin k x)

A + i B = \underset{k = 1}{\sum^{n}} e^{i k x} = \frac{e^{i (n + 1) x} - 1}{e^{i x} - 1}

A + i B = \frac{\cos (n + 1) x - 1 + i \sin (n + 1) x}{\cos x - 1 + i \sin x}

A + i B = \frac{[\cos (n + 1) x - 1 + i \sin (n + 1) x] (\cos x - 1 - i \sin x)}{(\cos x - 1 + i \sin x) (\cos x - 1 - i \sin x)}

A + i B = \frac{[\cos (n + 1) x - 1] (\cos x - 1) + \sin (n + 1) x \sin x + i {\sin (n + 1) x (\cos x - 1) - [\cos (n + 1) x - 1] \sin x}}{4 \sin^{2} \frac{x}{2}}

\Rightarrow A = \frac{[\cos (n + 1) x - 1] (\cos x - 1) + \sin (n + 1) x \sin x}{4 \sin^{2} \frac{x}{2}}

\Rightarrow A = \frac{\sin (n + \frac{1}{2}) x - \sin \frac{x}{2}}{2 \sin \frac{x}{2}}

B = \frac{\sin (n + 1) x (\cos x - 1) - [\cos (n + 1) x - 1] \sin x}{4 \sin^{2} \frac{x}{2}}

\Rightarrow B = \frac{\cos \frac{x}{2} - \cos (n + \frac{1}{2}) x}{2 \sin \frac{x}{2}}

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