# Trig equation

• Sep 24th 2011, 02:55 AM
elieh
Trig equation
I've found that in the first part of the question

$sin^5\theta = \frac{1}{16}(sin5\theta-5sin3\theta+10sin\theta)$

Now I have to solve $sin5\theta-5sin3\theta+6sin\theta = 0$ for theta

I can see straight away that 0 and multiples of pi are my solutions,
But I don't know how to do this problem in a methodical way.
I thought there would be substitution involved but I can't see how.

Help appreciated.
• Sep 24th 2011, 03:36 AM
Siron
Re: Trig equation
You can do it in a long way (I replace $\theta$ by $x$).
To solve:
$\sin(5x)-5\sin(3x)+6\sin(x)=0$

First we're going to expand $\sin(5x)$ and $\sin(3x)$ in terms of $\sin(x)$, therefore:
$\sin(5x)=\sin(2x+3x)=\sin(2x)\cdot \cos(3x)+\cos(2x)\cdot \sin(3x)=...=16\sin^5(x)-20\sin^3(x)+5\sin(x)$
$\sin(3x)=\sin(x+2x)=...=3\sin(x)-4\sin^3(x)$

Therefore the equation becomes:
$16\sin^5(x)-20\sin^3(x)+5\sin(x)-15\sin(x)+20\sin^3(x)+6\sin(x)=0$
$\Leftrightarrow 16\sin^5(x)-4\sin(x)=0$
$\Leftrightarrow 4\sin(x)[4\sin^4(x)-1]=0$
$\Leftrightarrow ...$
• Sep 24th 2011, 04:29 AM
elieh
Re: Trig equation
I dont' follow most of your steps
Which identities are you using?
• Sep 24th 2011, 04:35 AM
Siron
Re: Trig equation
Expand $\sin(5x)$ in terms of $\sin(x)$, do the same for $\sin(3x)$.
Can you do that?
Hint: rewrite $\sin(5x)=\sin(2x+3x)$ and $\sin(3x)=\sin(2x+x)$ and use the formula: $\sin(a+b)=\sin(a)\cdot \cos(b)+\sin(b)\cdot \cos(a)$ and also where necessary $\cos^2(x)=1-\sin^2(x)$.

It will take some time to do that.
• Sep 24th 2011, 05:03 AM
elieh
Re: Trig equation
Yeah I follow that,
Thanks!
• Sep 24th 2011, 05:16 AM
Siron
Re: Trig equation
You're welcome! Try to resolve the equation by your own calculations and offcourse take your time for doing that because in this case you have to pay attention with signs etc ... .
• Sep 24th 2011, 04:43 PM
SammyS
Re: Trig equation
$\sin^5\theta = \frac{1}{16}(\sin5\theta-5\sin3\theta+10\sin\theta)$

$16\sin^5\theta = \sin5\theta-5\sin3\theta+10\sin\theta$
$= (\sin5\theta-5\sin3\theta+6\sin\theta)+4\sin\theta$

$= (0)+4\sin\theta$

Therefore, $16\sin^5\theta = 4\sin\theta$ is equivalent to $\sin5\theta-5\sin3\theta+6\sin\theta=0\,.$