prove that; sin3(theta)=3sin(theta)-4sin^3(theta)

Question

anonymous · Answer

sin (1+2)theta

whpalmer4 · Answer

Is that supposed to be
$$\sin(3\theta) = 3\sin(\theta)-4\sin^3(\theta)$$
?

anonymous · Answer

yes

anonymous · Answer

sin(1+2)theta=sin1theta *cos2theta+sin2theta*cos1theta

whpalmer4 · Answer

Off the cuff, I would try something with the sum-difference formulas:

$$\sin(u\pm v) = \sin u \cos v \pm \cos u \sin v$$

You could use $u = 2\theta$ and $v = \theta$

whpalmer4 · Answer

Also $\cos(2u) = 2\cos^2u-1$

whpalmer4 · Answer

or $cos(2u) = 1-2\sin^2u$

anonymous · Answer

=sin1theta(cos^2theta-sin^2theta)+(2sin1theta*cos1 theta)*cos 1 theta

anonymous · Answer

ok what i dont understand is how sin 3(theta) can equal sin2(theta)cos(theta) +cos2(theta)sin(theta) and then = 2sin(theta)cos(theta)cos(theta)+(1-2(sin^2theta)sin(theta)?

anonymous · Answer

u just  sin 3 theta=sin (2+1)theta

anonymous · Answer

sin(1+2)theta=sin1theta *cos2theta+sin2theta*cos1theta

anonymous · Answer

ok i got that but the (1-2sin^2theta) bit is what i dont understand in that part

anonymous · Answer

use cos2theta=cos^2theta-sin^2theta

anonymous · Answer

sin^2theta=2sin1theta*costheta