Question

Given sin(alpha)=4/9 and pi/2<alpha<pi, find the exact value of sin(alpha/2)

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ANSWERED
asked 2021-03-09

Given \(\displaystyle \sin{{\left(\alpha\right)}}=\frac{4}{{9}}{\quad\text{and}\quad}\pi\text{/}{2}<\alpha<\pi\),
find the exact value of \(\displaystyle \sin{{\left(\alpha\text{/}{2}\right)}}.\)

Expert Answers (1)

2021-03-10

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

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