1)arcsin(cos 22п/5) 2)sin(arccos(-2/5))

Предыдущий
вопрос Следующий
вопрос

1) Let's find the value of arcsin(cos (22π/5)).
Since cos (22π/5) can be written as cos (4π + 2π/5), we can find the corresponding angle in the first quadrant by subtracting 4π from 22π/5.
22π/5 - 4π = 2π/5.
So, the angle in the first quadrant is 2π/5.
Therefore, arcsin(cos (22π/5)) = arcsin(cos (2π/5)) = 2π/5.

2) Let's find the value of sin(arccos(-2/5)).
Since arccos(-2/5) is the angle whose cosine is -2/5, and cosine is negative in the second and third quadrants, we need to find the reference angle in the first quadrant first.
Let's find the reference angle first:
cosθ = -2/5
sinθ = √(1 - cos^2(θ)) = √(1 - 4/25) = √(21/25) = √21/5
So, in the first quadrant, sin arccos(-2/5) = √21/5.
Since sin is negative in the third quadrant, sin(arccos(-2/5)) = -√21/5.