Well $\cos(3y)=\cos(y+2y)=\cos(y)\cos(2y)-\sin(y)\sin(2y)$. That's all I got. I've tried putting it in the equation but it doesn't seem to work out. How to solve this?
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You want to write $\cos{kx}$ as a polynomial entirely in $\cos{x}$, and leave $\sin$ out of it. – Andrew Dudzik Oct 04 '14 at 18:25
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how do I put Ant's edit back again? – MelanieJohansson Oct 04 '14 at 18:36
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Clearly $y\ne n\pi\implies\sin y\ne0$ for any integer $n$
Multiply either sides by $2\sin y,$ (this shows why $2\sin y$ was chosen as the multiplier)
and using Werner Formula $2\cos A\sin B=\sin(A+B)-\sin(A-B)$
to find $\displaystyle\sin6y=\sin y$
Now if $\sin x=\sin A,x=m\pi+(-1)^mA$ where $m$ is any integer
Don't forget to eliminate the extraneous root $y=m\pi$
lab bhattacharjee
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Well you need to get rid of $\cos(2y)=\cos(y+y)=\cos y\cos y-\sin y\sin y$.
$\cos(5y)=\cos(3y+2y)=\cos(3y)\cos(2y)-\sin(3y)\sin(2y)$
So by doing these kind of substitutions you should get an equation in $\cos^n(x)$.
Or you can skip and look here: http://www.analyzemath.com/trigonometry/trigonometric_formulas.html at point $15$.
Good luck!
