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 13 My Process: since sin⁡θ2=±1−cos⁡θ2 1−452⋅55=5−410=110

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

"Find sin⁡θ2 when sin⁡θ=35, and 0∘<θ<90∘ The answer..." was answered by our community

High School
Geometry
Trigonometry
Trigonometric equations and identities

Alvin Pugh

Answered question

2022-01-26

Find $\sin \frac{θ}{2}$ when $\sin θ = \frac{3}{5}$ , and $0^{\circ} < θ < 90^{\circ}$
The answer I'm currently getting is $\sqrt{\frac{1}{10}}$ but the answer must be either $\frac{\sqrt{10}}{10}, \frac{3 \sqrt{10}}{10} or \frac{1}{3}$
My Process:
since $\sin \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{2}}$
$\sqrt{\frac{1 - \frac{4}{5}}{2}} \cdot \sqrt{\frac{5}{5}} = \sqrt{\frac{5 - 4}{10}} = \sqrt{\frac{1}{10}}$

Answer & Explanation

Brenton Pennington

Beginner2022-01-27Added 15 answers

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

Johnny Cummings

Beginner2022-01-28Added 7 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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