Rearranging trigonometry formula How do I rearrange this trig formula to make y the subject? I am having trouble to take the $y$ out of the cosine.
$x = \cos(a+y)\times50$
Any help is appreciated.
 A: In plain english, $\sin(y) = ?$ is just asking, "what is the sine of the angle $y $?"
Similarly, I can ask the question "what is the angle whose sine is $y $?". This seems like a totally legit question, and we represent it as
$$\arcsin(y) = ? $$
The $\arcsin $ functions just asks for the angle that has the given sine. It can be seen as the function that undoes what $\sin $ did. Similarly for $\arccos $.
Simplify your expression to
$$\frac{x }{50} = \cos(y + a) $$
That is saying $\frac{x}{50}$ is the cosine of some angle $y+a $. So you ask yourself, what is the angle whose cosine is $\frac{x}{50} $? Well, from the previous sentence we already know it is $y+a $, and the question I worded can be written as 
$$\arccos\left(\frac{x}{50}\right) = y+a $$
Which leaves you with an equation you can manipulate in order to isolate $y $.
A: $$x = \cos(a+y)\times50$$
Divide both sides by $50$:
$$\frac{x}{50} =\frac{ \cos(a+y)\times50}{50}$$
$$\frac{x}{50}=\cos(a+y)$$
Now take the inverse function of cosinus:
$$a+y=cos^{-1}\left(\frac{x}{50}\right)$$
Subtract by $a$:
$$y+a-a=cos^{-1}\left(\frac{x}{50}\right)-a$$
This means that 
$$y=cos^{-1}\left(\frac{x}{50}\right)-a\:.$$
