Answer:
[tex]\frac{3\pi}{4}[/tex] or [tex]\frac{5\pi}{4}[/tex]
Step-by-step explanation:
We want to find the exact value of
[tex]arccos(-\frac{\sqrt{2} }{2})[/tex].
Let [tex]x=arccos(-\frac{\sqrt{2} }{2})[/tex].
This implies that,
Let [tex]cos(x)=-\frac{\sqrt{2} }{2}[/tex].
Since the cosine ratio is negative, it means the angle is in the second or third quadrant.
We first of all find
[tex]arccos(\frac{\sqrt{2} }{2})=\frac{\pi}{4}[/tex]
In the second quadrant,
[tex]x=\pi - \frac{\pi}{4} =\frac{3\pi}{4}[/tex]
In the third quadrant,
[tex]x=\pi +\frac{\pi}{4} =\frac{5\pi}{4}[/tex]
Any of them is correct on the interval,
[tex]0\le x\le 2\pi[/tex]
Answer:
arccos(-√2/2) = 3π/4 radians
Step-by-step explanation:
As we know Arccos(x)=y means we have to find the cos value of an "Angle y" for which the value is "x".
So as per question arccos(-√2/2) = arccos(-√2/(√2)²) OR (-1/√2)
Now we know that negative value of cos(x) lies in second quadrant.
If Cos x = 1/√2 then x = π/4
But the given value is (-1/√2) so cosx lies in second quadrant and the angle is = π-π/4 = 3π/4
So the arccos(-√2/2) = 3π/4.
