Solution assignment 06 Addition and subtraction formulas

Show:

$\sin^3(x)=\displaystyle\frac{3}{4}\sin(x)-\displaystyle\frac{1}{4}\sin(3x)$

We can rewrite the formula as:

$\sin(3x)=3\sin(x)-4\sin^3(x)$

For the left-hand side we have:

$\sin(3x)=\sin(2x+x)=\sin(2x)\cos(x)+\sin(x)\cos(2x)=$

$=2\sin(x)\cos(x)\cos(x)+\sin(x)(1-2sin^2(x))=$

$=2\sin(x)\cos^2(x)+\sin(x)-2\sin^3(x)=$

$=2\sin(x)(1-\sin^2(x))+\sin(x)-2\sin^3(x)=$

$=2\sin(x)-2\sin^3(x)+\sin(x)-2\sin^3(x)=$

$=3\sin(x)-4\sin^3(x)$