Question

Prove that $sinx+sin3x+......+sin(2n-1)x=\frac{si{n}^{2}nx}{sinx}$.

Answer 1

Let P(n) : $sinx+sin3x+....+sin(2n-1)x=\frac{si{n}^{2}nx}{sinx}$

For n = 1

$sinx=\frac{si{n}^{2}x}{sinx}$

$sinx=sinx$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$sinx+sin3x+......+sin(2k-1)x=\frac{si{n}^{2}kx}{sinx}$ .......(1)

We have to show that

$sinx+sin3x+....+sin(2k-1)x+sin(2k+1)x=\frac{si{n}^2(k+1)x}{sinx}$

Now,

$\{sinx+sin3x+....+sin(2k-1\left)x\right\}+sin(2k+1)x$

$=\frac{si{n}^{2}kx}{sinx}+\frac{sin(2k+1)x}{1}$

Using equation (i),

$=\frac{si{n}^{2}kx+sin(2k+1)sinx}{sinx}$

$=\frac{2si{n}^{2}kx+cos\left[\right(2k+1)x-x]-cos[2kx+x+x]}{2sinx}$

$=\frac{2si{n}^{2}kx+cos2kx-cos(2kx+2x)}{2sinx}$

$=\frac{1-cos2x(k+1)}{2sinx}$

$=\frac{si{n}^{2}x(k+1)}{sinx}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI.