# Half Sine problem

• Nov 24th 2012, 11:32 AM
M670
Half Sine problem
Given http://webwork.mathstat.concordia.ca...37aea14491.png and http://webwork.mathstat.concordia.ca...6cd2797d21.png, find the exact value of http://webwork.mathstat.concordia.ca...ca01dc2261.png.

$\displaystyle \cos(\alpha)=\frac{\sqrt15}{8}=1-\sin^2(\frac{\alpha}{2})$
$\displaystyle \sin^2(\frac{\alpha}{2})=1-\frac{\sqrt15}{8}=\frac{8-\sqrt15}{8}$ which then is $\displaystyle \sin(\frac{\alpha}{2})=\sqrt\frac{8-\sqrt15}{8}$

Where did I go wrong
• Nov 24th 2012, 11:53 AM
MarkFL
Re: Half Sine problem
I think you first want to find $\displaystyle \cos(\alpha)$ from:

$\displaystyle \cos^2(\alpha)=1-\left(\frac{7}{8} \right)^2=\frac{15}{64}$

Since $\displaystyle \alpha$ is a first quadrant angle, we have:

$\displaystyle \cos(\alpha)=\frac{\sqrt{15}}{8}$

Next, using the double-angle identity for sine and the half-angle identity for cosine, we find:

$\displaystyle \frac{7}{8}=2\sqrt{\frac{1+\frac{\sqrt{15}}{8}}{2} }\sin\left(\frac{\alpha}{2} \right)=\frac{1}{2}\sqrt{8+\sqrt{15}} \sin \left( \frac{ \alpha}{2} \right)$

Now solve for $\displaystyle \sin\left(\frac{\alpha}{2} \right)$
• Nov 24th 2012, 12:52 PM
M670
Re: Half Sine problem
Can you please give me the original formulas ? so I can better understand the substitution you have made?
• Nov 24th 2012, 01:24 PM
M670
Re: Half Sine problem
Should this not be $\displaystyle \sin\frac{\alpha}{2}=\sqrt\frac{1-\cos(u)}{2}$ then $\displaystyle \sin\frac{\alpha}{2}=\sqrt\frac{1-\frac{\sqrt15}{8}}{2}$
• Nov 24th 2012, 01:36 PM
Soroban
Re: Half Sine problem
Hello, M670!'

A couple of errors . . .

Quote:

$\displaystyle \text{Given: }\:\sin\alpha = \tfrac{7}{8},\;\text{ and }\,0 < \alpha < \tfrac{\pi}{2}$

$\displaystyle \text{Find the exact value of }\sin\tfrac{\alpha}{2}.$

$\displaystyle \cos^2\!\alpha \;=\;1-\sin^2\!\alpha \;=\;1 - \left(\tfrac{7}{8}\right)^2 \;=\;1-\tfrac{49}{64} \;=\;\tfrac{15}{64}$

. . $\displaystyle \text{Hence: }\:\cos\alpha \;=\:\frac{\sqrt{15}}{8}$

$\displaystyle \text{Identity: }\:\sin^2\tfrac{\alpha}{2} \:=\:\frac{1 - \cos\alpha}{{\color{red}2}} \;=\;\frac{1-\frac{\sqrt{15}}{8}}{2} \;=\;\frac{8-\sqrt{15}}{16}$

$\displaystyle \text{Therefore: }\:\sin\tfrac{\alpha}{2} \;=\;\sqrt{\frac{8-\sqrt{15}}{16}} \;=\; \frac{\sqrt{8-\sqrt{15}}}{4}$