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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.

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    We will have $a + y=\arccos \frac {x}{50} $. – 2017-01-21
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    Sorry @Rohan but i do not know what arccos means... – 2017-01-21
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    See [here](http://www.mathwords.com/c/cosine_inverse.htm) for more information. – 2017-01-21

2 Answers 2

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$$x = \cos(a+y)\times50$$

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

$$y=cos^{-1}\left(\frac{x}{50}\right)-a\:.$$

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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 $.

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    so arccos is basically the same thing as cos-1? – 2017-01-21
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    @kimchiboy03 yup – 2017-01-21