Half Sine problem

**M670** · November 24th 2012, 11:32 AM

Given

and

, find the exact value of

.
Note: You are not allowed to use decimals in your answer.

= .

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

Where did I go wrong

**MarkFL** · November 24th 2012, 11:53 AM

I think you first want to find $\cos(\alpha)$ from:

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

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

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

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

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

Now solve for $\sin\left(\frac{\alpha}{2} \right)$

**M670** · November 24th 2012, 12:52 PM

Can you please give me the original formulas ? so I can better understand the substitution you have made?

**M670** · November 24th 2012, 01:24 PM

Should this not be $\sin\frac{\alpha}{2}=\sqrt\frac{1-\cos(u)}{2}$ then $\sin\frac{\alpha}{2}=\sqrt\frac{1-\frac{\sqrt15}{8}}{2}$

**Soroban** · November 24th 2012, 01:36 PM

Hello, M670!'

A couple of errors . . .

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

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

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

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

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

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

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