Hi
I need to prove
sin^3 (x) = 1/4(3sinx -sin3x)
or
cos^3(x) = 1/4(3cosx+cos3x)
Either one will do because I should be able to solve the other one if I can find out how to solve one of them.
Hello, shadow85!
Prove: .$\displaystyle \sin^3x \:=\:\frac{3\sin x -\sin3x}{4}$
Consider: .$\displaystyle \sin3x \;=\;\sin(2x + x)$
. . . . . . . . . . . .$\displaystyle = \;\sin2x\cos x + \sin x\cos2x$
. . . . . . . . . . . .$\displaystyle = \;(2\sin x\cos x)\cos x + \sin x(2\cos^2\!x - 1)$
. . . . . . . . . . . .$\displaystyle = \;2\sin x\cos^2\!x + 2\sin x\cos^2\!x - \sin x$
. . . . . . . . . . . .$\displaystyle = \;4\sin x\cos^2\!x - \sin x$
. . . . . . . . . . . .$\displaystyle = \;4\sin x(1 - \sin^2\!x) - \sin x$
. . . . . . . . . . . .$\displaystyle = \;4\sin x - 4\sin^3\!x - \sin x $
. . . . . . . . . . . .$\displaystyle = \;3\sin x - 4\sin^3\!x$
The right side becomes: .$\displaystyle \frac{3\sin x - (3\sin x - 4\sin^3\!x)}{4} \;=\;\frac{4\sin^3\!x}{4} \;=\;\sin^3\!x$
$\displaystyle sin^3x$
= $\displaystyle sinx(sin^2x)$
= $\displaystyle sinx(1-cos^2x)$
= $\displaystyle sinx - sinxcosx^2x$
= $\displaystyle \frac{2sinx - 2sinxcos^2x}{2}$
= $\displaystyle sinx- \frac{sin2xcosx}{2}$
= $\displaystyle sinx - \frac{sin3x + sinx}{4}$
= $\displaystyle \frac {4sinx - sin3x -sinx}{4}$
= $\displaystyle \frac {3sinx- sin3x}{4}$
Here you go, I used these to help me out
$\displaystyle sinucosv = \frac {sin(u+v) + sin(u-v)}{2}$
$\displaystyle sin2u = 2sinucosu$
Hope this helps
Soroban's method is completely fine. He focused on one specific part of the RHS and modified it so that the entire RHS is equal to the LHS. If you really on insist,
$\displaystyle \begin{array}{rcl} \text{RHS} & = & \displaystyle\frac{3\sin x - {\color{red}\sin 3x}}{4} \\ & &\\ & = & \displaystyle\frac{3\sin x - {\color{red}\sin (2x + x)}}{4} \\ & & \\ & = & \displaystyle\frac{3\sin x - {\color{red}\big[\sin2x\cos x + \sin x\cos2x\big]}}{4} \\ & \vdots & \text{(Whatever Soroban did)} \\ & &\\ & = & \text{LHS} \end{array}$
Whatever is in red is exactly what Soroban did. Does this conform to what you wanted?
Also, the reason why he split sin(3x) to sin(2x + x) is so that you can use the formula: $\displaystyle \sin (A + B) = \sin A \cos B + \cos A \sin B$. We (normally) aren't given the formula of sin(3x) in terms of sin x.
$\displaystyle cosx - sin^2xcosx$
= $\displaystyle \frac{2cosx - (2sinxcosx)(sinx)}{2}$
= $\displaystyle cosx - \frac{(sin2x)(sinx)}{2}$
= $\displaystyle cosx - \frac{cosx - cos3x}{4}$
= $\displaystyle \frac{3cosx + cos3x}{4}$
Use this formula to help you
$\displaystyle (sinu)(sinv) = \frac{cos(u-v) - cos(u+v)}{2}$