Given sin⁡x=34 and π2&lt;x&lt;π, find sin⁡(x+π6) I have been trying to calculate this question sin⁡θ=34 π2&lt;θ&lt;π sin⁡(θ+π6)=sin⁡θ.cos⁡π6+cos⁡θ.sin⁡π6 =34.32+cos⁡θ.12 =3.3+4cos⁡θ8 The answer is: 3.3−78 However I do not know how to get to it.

You are on the right track. We have sin⁡(x+π6)=sin⁡xcos⁡π6+sin⁡π6cos⁡x=34⋅32+12cos⁡x and since sin⁡x=34=332+(7)2 we know that x&gt;π2 so it means that by Pythagoras, cos⁡x=−732+(7)2=−74 so the answer is 338−78

"Given sin⁡x=34 and π2<x<π, find sin⁡(x+π6) I have..." was answered by students like you

Madilyn Fitzgerald

Answered question

2022-01-30

Given $\sin x = \frac{3}{4}$ and $\frac{π}{2} < x < π$ , find $\sin (x + \frac{π}{6})$
I have been trying to calculate this question
$\sin θ = \frac{3}{4}$
$\frac{π}{2} < θ < π$
$\sin (θ + \frac{π}{6}) = \sin θ . \frac{\cos π}{6} + \cos θ . \frac{\sin π}{6}$
$= \frac{3}{4} . \frac{\sqrt{3}}{2} + \cos θ \frac{.1}{2}$
$= \frac{3 . \sqrt{3} + 4 \cos θ}{8}$
The answer is: $\frac{3 . \sqrt{3} - \sqrt{7}}{8}$
However I do not know how to get to it.

Answer & Explanation

stamptsk

Beginner2022-01-31Added 23 answers

You are on the right track.
We have

\sin (x + \frac{π}{6}) = \sin x \cos \frac{π}{6} + \sin \frac{π}{6} \cos x = \frac{3}{4} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cos x

and since

\sin x = \frac{3}{4} = \frac{3}{\sqrt{3^{2} + {(\sqrt{7})}^{2}}}

we know that

x > \frac{π}{2}

so it means that by Pythagoras,

\cos x = \frac{- \sqrt{7}}{\sqrt{3^{2} + {(\sqrt{7})}^{2}}} = - \frac{\sqrt{7}}{4}

so the answer is

\frac{3 \sqrt{3}}{8} - \frac{\sqrt{7}}{8}

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