Asked by Rachel
What is sin(cos^-1 a) when 0 < cos^-1 a < 90
Answers
Answered by
Steve
draw a triangle with leg a and hypotenuse 1.
As you can see, arccos(a) has sin sqrt(1-a^2)
As you can see, arccos(a) has sin sqrt(1-a^2)
Answered by
Steve
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)
Answered by
Rachel
Thank you. Could you explain in more detail please?
Answered by
Steve
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.