Question

Prove that sin 3x = 3 sinx - 4sin²x ​

Answer 1

Answer:

4sin×3÷3"(₹(₹(22(2)#)2)2(₹₹!₹

Answer 2

To prove: $sin3x=3sinx-4sin^{2} x$.

L.H.S: $sin3x=sin(2x+x)$

By applying formula: $sin(\alpha +\beta )=sin\alpha .cos\beta +cos\alpha .sin\beta$.

$=sin2x.cosx+cos2x.sinx$

We know that $sin2x=2sinx.cosx$ and $cos2x=1-2sin^{2} x$

$=(2sinx.cosx)\times cosx + (1-2sin^{2} x)\times sinx$

$=2sinx.cos^{2} x+sinx-2sin^{3} x$

We know that, $cos^{2} x=1-sin^{2} x$

$=2sinx(1-sin^{2} x)+sinx-2sin^{3} x$

$=2sinx-2sin^{3} x+sinx-2sin^{3} x$

$=2sinx+sinx-2sin^{3} x-2sin^{3} x$

$=3sinx-4sin^{3} x$ $=$ R.H.S.

Here, L.H.S is equal to R.H.S.

Hence, proved.