# How will you prove the formula sin3A=3sinA-4sin^3A using only the identity sin(A+B)=sinAcosB+cosAsinB?

Jul 23, 2016

The Given Identity

color(blue)(sin(A+B)=sinAcosB+cosAsinB........(1)

$\text{Inserting " color(red)(-B) " in place of B}$

color(red)(sin(A+B)=sinAcosB+cosAsinB........(2)

color(green)(sin(A+B)+sin(A-B)=2sinAcosB.......(3)

Putting 2A in place of B in (3)

color(blue)(sin(A+2A)+sin(A-2A)=2sinAcos2A

=>color(green)(sin3A-sinA=2sinAcos2A

=>color(green)(sin3A=2sinAcos2A+sinA................(4)

Putting (90-A) in place of B in (3)

color(blue)(sin(A+90-A)+sin(A-90+A)=2sinAcos(90-A)

=>color(blue)(sin(90)-sin(90-2A)=2sinAsinA

$\implies \textcolor{b l u e}{1 - \cos 2 A = 2 {\sin}^{2} A}$

$\implies \textcolor{b l u e}{\cos 2 A = 1 - 2 {\sin}^{2} A}$

$\text{Putting "color(red)(cos2A=1-2sin^2A) " in } \left(4\right)$

=>color(green)(sin3A=2sinAcos2A+sinA

=>color(green)(sin3A=2sinA(1-2sin^2A)+sinA.

$\therefore \textcolor{red}{\sin 3 A = 3 \sin A - 4 {\sin}^{3} A} .$

Proved