Answer to Question #103133 in Algebra for ea

Question

Express sin^5x
as a linear combination of sin kx and coskx, k belongs to integers.

Answer 1

We know

$sin3x=3sinx-4sin^3x$

$\Rightarrow sin^3x=\frac{3sinx-sin3x}{4}$

Also

$cos2x=1-2sin^2x$

$or \space sin^2x=\frac{1-cos2x}2$

We will use above two formula to write $sin^5x$ as linear combination of sine and cosine

$sin^5x$

$=sin^2x\cdot sin^3x$

$=\left(\frac{1-cos(2x)}{2}\right)\left(\frac{3sin(x)-sin(3x)}{4}\right)$

$=\frac{1}{8}\left(1-cos(2x)\right)\left(3sin(x)-sin(3x)\right)$

$=\frac{1}{8}\left(1\left(3sin(x)-sin(3x)\right)-cos(2x)\left(3sin(x)-sin(3x)\right)\right)$ [ multiplying

$=\frac{1}{8}\left(3sin(x)-sin(3x)-3cos(2x)sin(x)+sin(3x) cos(2x)\right)$ ...............(1)

This could be the answer but it can also further expanded as below using following formula

$sin(A+B)+sin(A-B)=2sinAcosB$

$Sin(A+B)-Sin(A-B)=2cosAcosB$

From (1)

$=\frac{3}{8}sin(x)-\frac{1}{8}sin(3x)-\frac{3}{8}cos(2x)sin(x)$ $+\frac{1}{8}sin(3x)cos(2x)$

$=\frac{3}{8}sin(x)-\frac{1}{8}sin(3x)-\frac{3}{16}*(2cos(2x)sin(x))$ $+\frac{1}{16}(2sin(3x)cos(2x))$

$=\frac{3}{8}sin(x)-\frac{1}{8}sin(3x)-\frac{3}{16}(sin(2x+x)$ $-sin(2x-x))+\frac{1}{16}(sin(3x+2x)+sin(3x-2x))$

$=\frac{3}{8}sin(x)-\frac{1}{8}sin(3x)-\frac{3}{16}sin(3x)$ $+\frac{3}{16}sin(x)+\frac{1}{16}sin(5x)+\frac{1}{16}sin(x)$

$=(\frac{3}{8}+\frac{3}{16}+\frac{1}{16})sin(x)+$ $(-\frac{1}{8}-\frac{3}{16})sin(3x)+\frac{1}{16}sin\left(5x\right)$

$=\frac{5}{8}sin(x)-\frac{5}{16}sin(3x)+\frac{1}{16}sin(5x)$