draw a triangle with leg a and hypotenuse 1.
As you can see, arccos(a) has sin sqrt(1-a^2)
What is sin(cos^-1 a) when 0 < cos^-1 a < 90
4 answers
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)
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.
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.