I have the problem$$\arccos\left(\cos \left(-\frac{7\pi}{10}\right)\right)$$But I am confused on it.
What I did: find reference angle, which is $\frac{3\pi}{10}$, and locate its quadrant, which is the first quadrant.
I am confused on what to do from here.
$\endgroup$ 02 Answers
$\begingroup$The arccos function arbitrarily restricts its range to $0 \leq \theta \leq \pi$. This restriction allows the arccos function to be well defined.
For example, if the arccos function allowed its range to include $-\pi/2 \leq \theta \leq 0$, then arccos$(1/\sqrt{2})$ would be ambiguously referring to either $\theta = (\pi/4)$ or $\theta = -(\pi/4)$.
So, since you are attempting to evaluate arccos$(\cos[-7\pi/10])$, the first thing to do is evaluate$A = \cos(-7\pi/10)$.
Noting that $-7\pi/10$ is in the 3rd quadrant, you have that $A < 0$. Therefore, the arccos$(A)$ must return a value in the 2nd quadrant.
Since, in general, $\cos(x) = \cos(-x),$ and since $(7\pi/10)$ is in the 2nd quadrant, the expression $~$ arccos$(\cos[-7\pi/10])~$ evaluates to $(7\pi/10)$.
$\endgroup$ 3 $\begingroup$$\cos(\theta)$ is - in Quadrants 2 and 3. $\frac{-7\pi}{10}$ will yield a negative value as the angle is in Quadrant 3. Therefore its co-arc is $-\cos\frac{3\pi}{10}$. Use this hint to find your reference angle. $$-\cos\frac{3\pi}{10} = \cos\frac{-7\pi}{10}$$.
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