# Homework question regarding inverse function with a cosine

I'm given $f(x)= 3-4\cos(x-2)$

I've gotten to $(-x+3)/4 = \cos(2)\cos(y)+\sin(s)\sin(y)$

But I can't get the $y$ out to create an inverse ...

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you mean $\cos(x-2)$? – Daniel Jul 13 '12 at 4:35
Gah, yes, sorry. – Daniel B. Jul 13 '12 at 4:37
Oh hey wait ... couldn't I take the cosine inverse of both sides, rather than using the difference identity to split it up, I mean. – Daniel B. Jul 13 '12 at 4:41
I think your idea is right – Daniel Jul 13 '12 at 4:49

Don't expand the $\cos(x-2)$. Solve the whole thing as you normally would, to get $x-2 = \cos^{-1}($something involving $f(x))$, and then just add 2 to both sides.

Spoiler:

$$\begin{eqnarray} f(x) & = & 3 - 4 \cos(x-2) \\ \frac14(3-f(x)) & = & \cos(x-2) \\ \cos^{-1}\left(\frac14(3-f(x))\right) &=& x-2 \\ 2+\cos^{-1}\left(\frac14(3-f(x))\right) &=& x\end{eqnarray}$$

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f(x) = 3 - 4cos(x-2)

we know that f(f^(-1)(x)) = x

write the function f(x) as an equation

y = 3 - 4cos(x-2)

solve the equation for x

x = [arccos((3-y)/4) + 2]

now write the f^(-1)(y)

f^(-1)(y) = [arccos((3-y)/4) + 2]

replace y with x in the equation above

i.e. f^(-1)(x) = [arccos((3-x)/4) + 2]

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