How to evaluate \sin^{-1}(\sqrt{2}\sin\theta)+\sin^{-1}(\sqrt{\cos\theta})

Davian Lawson

Davian Lawson

Answered question

2022-04-20

How to evaluate
sin1(2sinθ)+sin1(cosθ)

Answer & Explanation

Ronnie Porter

Ronnie Porter

Beginner2022-04-21Added 12 answers

Set a=2sinθ and b=cos2θ. It follows that a2+b2=2sin2θ+cos2θ=1.
If α=sin1a where α[π2,π2], we get sin(α)=a, and since b0 we have b=1a2=cos(α)=sin(π2α)
If a0, we get a[0,π2] and thus sin1b=π2α which gives us sin1a+sin1b=π2
On the other hand, if a<0, we have α[π2,0) and so sin1b=π+α so sin1a+sin1b has no fixed value.

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