What is sin(cos^-1 a) when 0 < cos^-1 a < 90

draw a triangle with leg a and hypotenuse 1.

As you can see, arccos(a) has sin sqrt(1-a^2)

or, fall back on the good old identity

sin^2(arccos(a)) + cos^2(arccos(a)) = 1
but cos(arccos(a)) = a, so
sin^2(arccos(a)) + a^2 = 1
sin^2(arccos(a) = 1-a^2
sin(arccos(a)) = sqrt(1-a^2)

Thank you. Could you explain in more detail please?

if you draw a triangle with one leg a and the hypotenuse 1, then the other leg is √(1-a^2)

so, if the anglarccos(a) is θ, the adjacent/hypotenuse is cosθ = a/1 = a
opposite/hypotenuse is sinθ = √(1-a^2)

so, going back to θ

sin^2θ + cos^2θ = 1
if that isn't clear, you have some major catching up to do!
but, we know that cosθ = a, so
sin^2θ + a^2 = 1
sin^2θ = 1-a^2
sinθ = √(1-a^2)

I think that's about as clear as I can make it.

To find the value of sin(cos^-1 a) when 0 < cos^-1 a < 90, we can use the Pythagorean identity. The Pythagorean identity states that sin^2(x) + cos^2(x) = 1 for any angle x.

In this case, we have cos^-1 a as our angle, which is an acute angle since it is between 0 and 90 degrees. This means that sin^2(cos^-1 a) + cos^2(cos^-1 a) = 1.

Let's use the trigonometric identity cos(cos^-1 a) = a and sin(cos^-1 a) = √(1 - a^2) to solve for sin(cos^-1 a).

We know that cos(cos^-1 a) = a, so substituting this into our equation, we have sin^2(cos^-1 a) + a^2 = 1.

Rearranging the equation, we get sin^2(cos^-1 a) = 1 - a^2.

Now, taking the square root of both sides, we have sin(cos^-1 a) = √(1 - a^2).

Therefore, sin(cos^-1 a) when 0 < cos^-1 a < 90 is √(1 - a^2).