How do you find the exact value of #arccos(-1/sqrt(2))#?
First, it would be helpful to rationalize
Arccos is asking for the ANGLE with a cosine of the given value.
The range of arccos is between zero and
According to the unit circle, the angle in the second quadrant (between
cos x = - 1/(sqrt2) = - sqrt2/2
On the trig unit circle, there are 2 arcs that have the same cos value:
Check with calculator:
First we should understand what the question is about.
It means that, we need to find an angle, when it is inside a cosine function gives
Let's rationalize it
Now let's find out the angle using the unit circle
The angle is