Question

How do you calculate $\arccos \left( -\dfrac{\sqrt{3}}{2} \right)?$

Answer 1

Hint: We will equate the given inverse trigonometric value to a variable. Then we will find the corresponding angle. We will subtract the angle from the upper limit of the range of the inverse cosine function to find the principal value.

Complete step by step solution:
Consider the given inverse trigonometric value, $\arccos \left( -\dfrac{\sqrt{3}}{2} \right).$
The term $\arccos $ is an inverse trigonometric function which denotes the inverse cosine function. So, we are asked to find the principal value of the given inverse trigonometric function.

Let us suppose that $x=\arccos \left( -\dfrac{\sqrt{3}}{2} \right).$
Now after a rearrangement we will get $\cos x=-\dfrac{\sqrt{3}}{2}.$
Now let us consider the trigonometric cosine function $\cos y=\dfrac{\sqrt{3}}{2}.$
We know that the value of $y$ is $\dfrac{\pi }{6}$ when the value of $\cos y=\dfrac{\sqrt{3}}{2}.$
That is, $y=\dfrac{\pi }{6}.$
However, the value we have to deal with possesses a negative sign.
That is $-\dfrac{\sqrt{3}}{2}.$
So, we need to think of the situations where the value of the cosine function is negative.
There are only two situations where the cosine function is negative. That is when the function lies either in the second quadrant or in the third quadrant.
Here we are dealing with the inverse cosine function.
Also, we know that the range of the inverse cosine function is the set $\left[ 0,\pi \right].$
So, let us confirm that the function lies in the second quadrant.
So, we subtract $\dfrac{\pi }{6}$ from $\pi $ to get the exact value of $\arccos \left( -\dfrac{\sqrt{3}}{2} \right).$
Now we will get $\cos \left( \pi -\dfrac{\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}.$
Hence the value of $\arccos \left( -\dfrac{\sqrt{3}}{2} \right)=\pi -\dfrac{\pi }{6}.$

Note: Remember that $\cos \left( -\dfrac{\pi }{6} \right)=\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}.$ Also, remember that $\arccos \left( -\dfrac{\sqrt{3}}{2} \right)\ne -\dfrac{\pi }{6}.$ We should remember that the range of $\arccos x$ is $\left[ 0,\pi \right].$ The range of $\arcsin x$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and the range of the $\arctan x$ is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right).$