Given that $\dfrac{3\pi}{2} < z < 2\pi$
$x = arccos(sin(z))$
Given different values for z (which are angles on the unit circle) how would I write the results in these two forms, where C is a constant?:
a ) $x = z + C$
b ) $x = -z + C$
Say angle $z = \dfrac{7\pi}{4}$ and $z =\dfrac{ 5\pi}{3}$
I've drawn the unit circle and everything..
For example, when $z = \dfrac{7\pi}{4}$, I know $x$ would $= arccos(sin(z)).. z = arccos(-\sqrt{2}/2)$
I also know that $cos = ({\sqrt{2}}/2)$ at $\dfrac{5\pi}{4}$ and $\dfrac{3\pi}{4}$
But I'm not sure how to represent those in one of the forms given.
Let's do the first one, where $z=\frac{7\pi}{4}$. The angle $z$ is in the fourth quadrant. Its sine is $-\frac{1}{\sqrt{2}}$. Now take the $\arccos$ of this number. since the number is negative, the $\arccos$ of the number, that is, $x$, is $\frac{3\pi}{4}$.
Now we are asked to find a number $C$ such that $x=z+C$. Since $x=\frac{3\pi}{4}$ and $z=\frac{7\pi}{4}$, we get $C=\frac{3\pi}{4}-\frac{7\pi}{4}$. So $C=-\pi$.
The second question asks for $x=-z+C$. So $C=\frac{3\pi}{4}+\frac{7\pi}{4}=\frac{5\pi}{2}$.
The other problrm can be done using similar reasoning.