Given that -\frac{\pi}{2}\leq\arcsin x\leq\frac{\pi}{2}, find the exact of \cos[\arcsin(-\frac{1}{3})]

tressudasikx

tressudasikx

Answered question

2022-02-27

Given that π2arcsinxπ2, find the exact of cos[arcsin(13)]

Answer & Explanation

jaewonlee0217fyv

jaewonlee0217fyv

Beginner2022-02-28Added 5 answers

Use, cos2θ+sin2θ and also sin(arcsin(y))=y to find that
cos2(arcsin(13))=1(13)2=89
Now observe that π2arcsin(13))π2, so
cos(arcsin(13))0
So it is the positive root of the above quadratic equation:
cos(arcsin(13))=89=223

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