Given sin⁡(α)=49andπ/2&lt;α&lt;π, find the exact value of sin⁡(α/2).

Step 1 π2&lt;α&lt;π π22&lt;α2&lt;π2 π4&lt;α2&lt;π2 So, α2 in forst quadrant. In first quadrant sin is positive. Step 2 sin⁡(α2)=1−cos⁡α2 So, we have to find cos alpha from sin⁡α π2&lt;α&lt;π is second quadrant. There cos⁡α is negative cos⁡α=−1−sin2α=−1−(49)2=−1−1681=−659 Step 3 Plug cos⁡α=−659 in the formula sin⁡(α2)=1−cos⁡α2 sin⁡(α2) =1−(−659)2 =1+6592 =9+6518 Multiply numerator and denominator by 2 =2(9+65)18(2) =18+26536 =13+2135+56 =(13+5)26 =13+56 Answer: =13+56

"Given sin⁡(α)=49andπ/2<α<π, find the exact value of sin⁡(α/2)." was answered by students like you

High School
Other

Reeves

Answered question

2021-03-09

Given $\sin (α) = \frac{4}{9} and π / 2 < α < π$ ,
find the exact value of $\sin (α / 2) .$

Answer & Explanation

davonliefI

Skilled2021-03-10Added 79 answers

Step 1
$\frac{π}{2} < α < π$
$\frac{\frac{π}{2}}{2} < \frac{α}{2} < \frac{π}{2}$
$\frac{π}{4} < \frac{α}{2} < \frac{π}{2}$
So, $\frac{α}{2}$ in forst quadrant.
In first quadrant sin is positive.
Step 2 $\sin (\frac{α}{2}) = \sqrt{\frac{1 - \cos α}{2}}$
So, we have to find cos alpha from $\sin α$
$\frac{π}{2} < α < π$ is second quadrant.
There $\cos α$ is negative
$\cos α = - \sqrt{1 - \sin^{2} α} = - \sqrt{1 - {(\frac{4}{9})}^{2}} = - \sqrt{1 - \frac{16}{81}} = - \frac{\sqrt{65}}{9}$
Step 3
Plug $\cos α = - \frac{\sqrt{65}}{9}$ in the formula
$\sin (\frac{α}{2}) = \sqrt{\frac{1 - \cos α}{2}}$
$\sin (\frac{α}{2})$
$= \sqrt{\frac{1 - (- \frac{\sqrt{65}}{9})}{2}}$
$= \sqrt{\frac{1 + \frac{\sqrt{65}}{9}}{2}}$
$= \sqrt{\frac{9 + \sqrt{65}}{18}}$
Multiply numerator and denominator by 2
$= \sqrt{\frac{2 (9 + \sqrt{65})}{18 (2)}}$
$= \sqrt{\frac{18 + 2 \sqrt{65}}{36}}$
$= \sqrt{13 + 2 \sqrt{13} \frac{\sqrt{5 + 5}}{6}}$
$= \frac{\sqrt{{(\sqrt{13} + \sqrt{5})}^{2}}}{6}$
$= \frac{\sqrt{13} + \sqrt{5}}{6}$
Answer: $= \frac{\sqrt{13} + \sqrt{5}}{6}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Other

Didn't find what you were looking for?