# Thread: Inverse trig functions involving the unit circle?

1. ## Inverse trig functions involving the unit circle?

How do I do these problems?

1) cos ^ -1 ( cos (7 pi/4) )

2) sin ^ -1 ( - sq. root 3 /2 )

The answer to the first problem is supposed to be pi/4 and 7 pi/4
but I don't get how to do these? I know our answers are to be in terms of the unit circle, so if you can explain it in detail please do so, and were not allowed to use a calculator to solve these

I have no clue on how to do these inverse trig functions, I asked on yahoo answers and people weren't explaining well enough

2. Originally Posted by realintegerz
How do I do these problems?

1) cos ^ -1 ( cos (7 pi/4) )

2) sin ^ -1 ( - sq. root 3 /2 )

The answer to the first problem is supposed to be pi/4 and 7 pi/4
but I don't get how to do these? I know our answers are to be in terms of the unit circle, so if you can explain it in detail please do so, and were not allowed to use a calculator to solve these

I have no clue on how to do these inverse trig functions, I asked on yahoo answers and people weren't explaining well enough
From the definition of "inverse" $\displaystyle cos^{-1}(cos(7\pi/4))= 7\pi/4$ is the obvious solution (f^{-1} is defined by $\displaystyle f^{-1}(f(x))= f(f^{-1}(x))= x$). Since cosine is the x coordinate of a point on the unit circle, mark the point at angle $\displaystyle 7\pi/4$ which is in the 4th quadrant, at angle $\displaystyle 2\pi- 7\pi/4= \pi/4$ below the x-axis, and draw a vertical line there. You should be able to see by symmetry that the other point at which the line crosses the unit circle is at $\displaystyle \pi/4$ above the x-axis.