1. ## trig identity

prove that sin(3x)=3sinx-4sin^3x

2. Use the fact $\sin3x=\sin(x+2x)$

Then use $\sin(\alpha+\beta)$ formula.

Of course remember $\sin2x$ and $\cos2x$

3. Hello, polymerase!

Here it is . . . in baby-steps.

Prove that: . $\sin(3x)\:=\:3\sin x - 4\sin^3x$
We're expected to know these identities:

. . $\begin{array}{ccc}\sin(A + B) & = & \sin A\cos B + \sin B\cos A \\ \sin2x & = & 2\sin x\cos x \\ \cos2x & = & 2\cos^2x - 1 \\ \cos^2\!x & = & 1 - \sin^2\!x\end{array}$

$\sin3x \:=\:\sin(2x+x)$

. . . . $=\;\sin2x\cos x + \sin x\cos2x$

. . . . $= \;(2\sin x\cos x)\!\cdot\cos x + \sin x(2\cos^2\!x-1)$

. . . . $= \;2\sin x\cos^2\!x + 2\sin x\cos^2\!x - \sin x$

. . . . $= \;4\sin x\cos^2\!x - \sin x$

. . . . $= \;4\sin x(1-\sin^2\!x) - \sin x$

. . . . $= \;4\sin x - 4\sin^3\!x - \sin x$

. . . . $= \;3\sin x - 4\sin^3\!x$

4. Another approach:

$\cos 3x+i\sin 3x=(\cos x+i\sin x)^3=$
$=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x$
Identifying the real parts and the imaginary parts we have
$\cos 3x=\cos^3x-3\cos x\sin^2x=\cos^3x-3\cos x(1-\cos^2x)=4\cos^3x-3\cos x$
$\sin 3x=3\cos^2x\sin x-\sin^3x=3(1-\sin^2x)\sin x-\sin^3x=3\sin x-4\sin^3x$

5. Originally Posted by red_dog
Another approach:

$\cos 3x+i\sin 3x=(\cos x+i\sin x)^3=$
$=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x$
Identifying the real parts and the imaginary parts we have
$\cos 3x=\cos^3x-3\cos x\sin^2x=\cos^3x-3\cos x(1-\cos^2x)=4\cos^3x-3\cos x$
$\sin 3x=3\cos^2x\sin x-\sin^3x=3(1-\sin^2x)\sin x-\sin^3x=3\sin x-4\sin^3x$
nice! how do you come up with this stuff?!

6. Originally Posted by Jhevon
nice! how do you come up with this stuff?!
This is actually a well known trick.

To find $\sin nx \mbox{ and }\cos nx$.
Use de Moiver's Theorem.

So, $(\cos x + i \sin x)^n = \cos nx + i \sin nx$.
Apply Binomial Expansion on RHS.