Question

Question #6cb94

Answer 1

By using Demoivre's theorem multiply the trig expression out and extract the imaginary part.

Explanation:

we shall use `(costheta+isintheta)^n=cosntheta=isinntheta`

Since we want `sin4theta` we shall let `n=4`

so `sin4theta = Im((costheta+isintheta)^4)`
since `(costheta+isintheta)^4=cos4theta+isin4theta`
expand `(costheta+isintheta)^4`

# (costheta+isintheta)^4= cos^4theta +4(cos^3theta(isintheta)) +6(cos^2theta(isintheta)^2)+4(costheta(isintheta)^3)+(isintheta)^4 #

# (costheta+isintheta)^4= cos^4theta +4i(cos^3thetasintheta) -6(cos^2thetasin^2theta)-4i(costhetasin^3theta))+sin^4theta #
now extract the imaginary part

`sin4theta = Im((costheta+isintheta)^4)`

`sin4theta=4cos^3thetasintheta-4costhetasin^3theta`

`sin4theta=4(cos^3thetasintheta-costhetasin^3theta)`

as required

Answer 2

Prove sin 4t

Explanation:

Use trig identity: sin (a + b) = sin a.cos b + sin b.cos a
sin 4t = sin (2t + 2t) = sin 2t. cos 2t + cos 2t.sin 2t = 2sin 2t.cos 2t =
Since,
sin 2t = 2sin t.cos t
`cos 2t = cos^2 t - sin^2 t`,
There for:
`sin 4t = 4sin t.cos t(cos^2 t - sin^2 t)`
`sin 4t = 4(sin t.cos^3 t - cos t.sin^3 t)`