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How can I prove this equality please? equality

Edit: MathJax version:

$$ \sum_{k=0}^{n}\sin (kx)=\frac{\sin\left(\frac{n+1}{2}x\right)\sin\left(\frac{n}{2}x\right)}{\sin\left(\frac{x}{2}\right)} $$ $$ \sum_{k=0}^{n}\cos (kx)=\frac{\sin\left(\frac{n+1}{2}x\right)\cos\left(\frac{n}{2}x\right)}{\sin\left(\frac{x}{2}\right)} $$

Nathanael Skrepek
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Elisabeth
  • 131

2 Answers2

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Hint:

The left-hand sides are respectively the imaginary and real parts of the partial sum of the geometric series $$\sum_{k=0}^n\mathrm e^{ikx}=\frac{-1}{e^{ix}-1}.$$ Factor out $\;\mathrm e^{i\tfrac{(n+1)x}2}$ in the numerator,$\;\mathrm e^{i\tfrac{x}2}$ from the denominator, and use Euler's formulæ:

\begin{align} \sum_{k=0}^n\mathrm e^{ikx}&=\frac{\mathrm e^{\tfrac{i(n+1)x}2}}{\mathrm e^{i\tfrac{x}2}}\frac{\mathrm e^{\tfrac{i(n+1)x}2}-\mathrm e^{-\tfrac{i(n+1)x}2}}{\mathrm e^{\tfrac{ix}2}-\mathrm e^{-\tfrac{ix}2}}=\mathrm e^{i\tfrac{nx}2}\frac{2i\sin\dfrac{(n+1)x}2}{2i\sin\dfrac x2}\\\ &=\Bigl(\cos\frac{nx}2+i\sin\frac{nx}2\Bigr)\frac{\sin\dfrac{(n+1)x}2}{\sin\dfrac x2}. \end{align}

Bernard
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Hint: multiply by $$\frac{\sin \frac x2}{\sin \frac x2}$$ then use product to sum formulas $$\sum_{k=1}^{n}\sin kx=\sin x +\sin 2x +...+\sin nx\\\to \frac{\sin \frac x2}{\sin \frac x2}\times (\sin x +\sin 2x +...+\sin nx)$$

Khosrotash
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