Find sin⁡θ2 when sin⁡θ=35, and 0∘&lt;θ&lt;90∘The answer I'm currently getting is 110 but the answer must be either 1010,31010 or 13My Process:since sin⁡θ2=±1−cos⁡θ21−452⋅55=5−410=110

I had forgotten that, 110=110=110⋅(1010)=1010So, there is my answer

Expert Answer to "Find sin⁡θ2 when sin⁡θ=35, and 0∘<θ<90∘The answer I'm..."

Julian Peck

2022-04-15

Find

\sin \frac{θ}{2}

when

\sin θ = \frac{35}{}

, and

0^{\circ} < θ < 90^{\circ}

The answer I'm currently getting is

\sqrt{\frac{1}{10}}

but the answer must be either

\frac{\sqrt{10}}{10}, \frac{3 \sqrt{10}}{10} or \frac{13}{}

My Process:
since

\sin \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{2}}

\sqrt{\frac{1 - \frac{4}{5}}{2}} \cdot \sqrt{\frac{5}{5}} = \sqrt{\frac{5 - 4}{10}} = \sqrt{\frac{1}{10}}

cadhail6n1t

Beginner2022-04-16Added 14 answers

\frac{\sqrt{10}}{10} = \frac{\sqrt{10}}{\sqrt{10} \times \sqrt{10}} = \frac{1}{\sqrt{10}}

webhui2v2

Beginner2022-04-17Added 10 answers

I had forgotten that,

\sqrt{\frac{1}{10}} = \frac{\sqrt{1}}{\sqrt{10}} = \frac{1}{\sqrt{10}} \cdot (\frac{\sqrt{10}}{\sqrt{10}}) = \frac{\sqrt{10}}{10}

So, there is my answer

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