To find the cotangent of the angle 2π - 3arcsin(√2/2), we can first simplify the expression inside the inverse sine function.
We know that sin(π/4) = √2/2, so we can rewrite √2/2 as sin(π/4).
Now, we have:2π - 3arcsin(sin(π/4))
Since the sine function is periodic with a period of 2π, arcsin(sin(π/4)) = π/4. Thus, we have:2π - 3(π/4) = 2π - 3π/4 = 8π/4 - 3π/4 = 5π/4
Now, we can find the cotangent of 5π/4:cot(5π/4) = cot(π/4) = 1
Therefore, cot(2π - 3arcsin(√2/2)) = 1.
To find the cotangent of the angle 2π - 3arcsin(√2/2), we can first simplify the expression inside the inverse sine function.
We know that sin(π/4) = √2/2, so we can rewrite √2/2 as sin(π/4).
Now, we have:
2π - 3arcsin(sin(π/4))
Since the sine function is periodic with a period of 2π, arcsin(sin(π/4)) = π/4. Thus, we have:
2π - 3(π/4) = 2π - 3π/4 = 8π/4 - 3π/4 = 5π/4
Now, we can find the cotangent of 5π/4:
cot(5π/4) = cot(π/4) = 1
Therefore, cot(2π - 3arcsin(√2/2)) = 1.