Find the exact value:
1.) cos^-1(sin(pi/2))
2.) arccos(cos(7pi/2))
1. cos^-1(sin(π/2)
we know sin π/2 = 1
so cos^-1 (sin π/2)
= cos^-1 (1)
= 0
(because cos 0 = 1 )
2. this one is easy, since arccos is the same as cos^-1
the answer is obviously 7π/2
However, they might just want a reduced answer
since cos 7π/2 = cos 3π/2
arccos(cos(7π/2) )
= arccos(cos (3π/2))
=3π/2
To find the exact values of these trigonometric expressions, we need to use the unit circle and specifically look at the angle(s) involved.
1.) The expression cos^⁻¹(sin(π/2)) asks us to find the cosine of an angle whose sine is (sin(π/2)). In other words, we need to find the angle on the unit circle whose sine value is 1.
The unit circle is a circle with a radius of 1 centered at the origin (0, 0) of the coordinate plane. The angle π/2 corresponds to the point (0, 1) on the unit circle.
Since we are looking for the cosine of this angle, we need to find the x-coordinate of the point (0, 1). Since the x-coordinate at this point is 0, the cosine of π/2 is 0.
So, cos^⁻¹(sin(π/2)) = cos^⁻¹(1) = 0.
2.) The expression arccos(cos(7π/2)) asks us to find the angle (in radians) whose cosine is the same as that of (cos(7π/2)).
The angle 7π/2 is equivalent to rotating 3 and a half times around the unit circle.
The cosine function has a period of 2π, meaning that the cosine values repeat after an interval of 2π.
Since 7π/2 exceeds the period of the cosine function, we need to determine the equivalent angle within one period.
Since 7π/2 is the same as 3 and a half rotations counterclockwise around the unit circle, we can subtract 2π from it to find the equivalent angle within one period:
7π/2 - 2π = 7π/2 - 4π/2 = 3π/2
Now, we need to find the angle whose cosine is the same as that of (cos(3π/2)).
If we look at the point (0, -1) on the unit circle, we see that its cosine value is -1.
Therefore, arccos(cos(7π/2)) = arccos(cos(3π/2)) = arccos(-1) = π.