# Thread: help with proving sin^3(x)

1. ## help with proving sin^3(x)

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.

Prove: . $\sin^3x \:=\:\frac{3\sin x -\sin3x}{4}$

Consider: . $\sin3x \;=\;\sin(2x + x)$

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

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

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

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

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

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

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

The right side becomes: . $\frac{3\sin x - (3\sin x - 4\sin^3\!x)}{4} \;=\;\frac{4\sin^3\!x}{4} \;=\;\sin^3\!x$

3. Hi im not sure why you started of with sin(3x) = sin(2x+x). What I was hoping you would do is start with sin^3(x) = sin^2(x) Sin(x). And then go from there, is it possible this way? because what I am trying to prove is sin cubed x, not sin3x.

4. I proved that the right side equals the left side.
That suffices as a proof.

Going the other way is very difficult.
Did you try it?

5. Originally Posted by Soroban
I proved that the right side equals the left side.
That suffices as a proof.

Going the other way is very difficult.
Did you try it?
Lol I tried the other way and yep it a pain.

6. I have started but always cant get down to the solution. But I need to do it this way because it says prove using trig identities that LHS = RHS as i write above.

If anyone can do it that way, id b happy to donate.

7. $sin^3x$

= $sinx(sin^2x)$

= $sinx(1-cos^2x)$

= $sinx - sinxcosx^2x$

= $\frac{2sinx - 2sinxcos^2x}{2}$

= $sinx- \frac{sin2xcosx}{2}$

= $sinx - \frac{sin3x + sinx}{4}$

= $\frac {4sinx - sin3x -sinx}{4}$

= $\frac {3sinx- sin3x}{4}$

Here you go, I used these to help me out

$sinucosv = \frac {sin(u+v) + sin(u-v)}{2}$

$sin2u = 2sinucosu$

Hope this helps

8. 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,

$\begin{array}{rcl} \text{RHS} & = & \displaystyle\frac{3\sin x - {\color{red}\sin 3x}}{4} \\ & &\\ & = & \displaystyle\frac{3\sin x - {\color{red}\sin (2x + x)}}{4} \\ & & \\ & = & \displaystyle\frac{3\sin x - {\color{red}\big[\sin2x\cos x + \sin x\cos2x\big]}}{4} \\ & \vdots & \text{(Whatever Soroban did)} \\ & &\\ & = & \text{LHS} \end{array}$

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: $\sin (A + B) = \sin A \cos B + \cos A \sin B$. We (normally) aren't given the formula of sin(3x) in terms of sin x.

9. Another way to prove the LHS = RHS

$LHS - RHS = 0$

10. Hi, thanks 11rdc11. I think that should do it, what I am after. You guys are so krazy at maths, lol. I havent done maths for a whole year then this was one of my assignment questions, lol.

11. OK with the cos^3(x).

I started with
=cos^2x*cosx
=(1-sin^2(x))*cosx
=cosx-cosxsin^2x
=(2cosx-2cosxsin^2x)/2

but I get stuck here.

I know in the sin^3x it was easier because sin2x=2sinxcosx

but with this cos^3x im stuck, can anyone help from here?

12. What are you trying to show?

13. Originally Posted by o_O
What are you trying to show?
cos^3(x) = 1/4(3cosx+cos3x)

Hmm give me a few and ill try to get you an answer, just kind of busy packing up getting ready to evacuate again thanks to another hurricane.

It may be easier making the right hand side equal the left like soroban and o O said

14. $cosx - sin^2xcosx$

= $\frac{2cosx - (2sinxcosx)(sinx)}{2}$

= $cosx - \frac{(sin2x)(sinx)}{2}$

= $cosx - \frac{cosx - cos3x}{4}$

= $\frac{3cosx + cos3x}{4}$

$(sinu)(sinv) = \frac{cos(u-v) - cos(u+v)}{2}$

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# Sin^3 x=1/4(3sinx - sin 3x)

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