[calc 1] How do you simplify/get cos(arccsc(x)) to √(x^2-1)/x ?

Question

Thank you in advance!!!!!

Answer 1

Another way is to create a right triangle where csc(x) is equal to x. To do this, make θ one of the non-right angles, the hypotenuse x, and the side opposite θ equal to 1. This makes csc(θ) = x, and therefore the csc^-1(x) = θ.

Then use the Pythagorean Theorem to get the remaining side (which will be the adjacent to θ). This side will be √(x² - 1). Then cos(csc^-1(x)) = cos(θ) = √(x² - 1)/x.

Answer 2

sin(θ) = 1/csc(θ)

cos(θ) = √(1 - sin²(θ) = √(1 - 1/csc²(θ) = √((csc²(θ) -1)/csc²(θ)) = √(csc²(θ) -1)/csc(θ)

Then, substituting x for csc(θ), we get √(x^2-1)/x

Thus, if the csc of an angle is x, the cos of the angle is √(x^2-1)/x

cos(arccsc(x)) is the cos of an angle whose csc is x

Then, the cos of an angle whose csc is x is the cos of an angle whose cos is √(x^2-1)/x, which is √(x^2-1)/x.