Question

How do you prove `sin 3 theta = 3 sin theta - 4 sin^3 theta`?

Answer 1

see explanation

Explanation:

Express the left hand side as

`sin3theta=sin(theta+2theta)`

now expand the right side of this equation using `color(blue)"Addition formula"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))`

`rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)`

`color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))`

The right hand side is expressed only in terms of `sintheta's`

so we use `cos2theta=1-2sin^2theta........(1)`

`color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)`

Replace `cos2theta" and " sin2theta` by the expansions (1) and (2)
into (A)

`sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)`

and expanding brackets gives.

`sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)`

`color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))`

Replace `cos^2theta=1-sin^2theta" into (B)"`

`rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)`

and expanding 2nd bracket gives.

`sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta`

Finally, collecting like terms.

`sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"`