Exact value of sin(5pi/12) + sin(pi/12)

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smsmith

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I could use some help if anyone has any input. The first half is what is causing my problem. sin(5pi/12) + sin(pi/12). I have already figured sin (pi/12) using difference formula. which results in the square root of 6 - the square root of 2 divided by 4.

I am having problems with the first term. Is it 2pi/3 -Pi/4? And if so what is the sin and cosine of 2pi/3 ? Any input would be appreciated.
 
Hello, smsmith!

Evaluate: \(\displaystyle \,\sin\left(\frac{5\pi}{12}\right)\,+\,\sin\left\frac{\pi}{12}\right)\)

I have already figured \(\displaystyle \sin\left(\frac{\pi}{12}\right)\) using difference formula.
which results in: \(\displaystyle \:\frac{\sqrt{6}\,-\,\sqrt{2}}{4}\;\;\) . . . Right!

I am having problems with the first term. .Is it \(\displaystyle \frac{2\pi}{3}\,-\,\frac{\pi}{4}\) ? . . . well, yes
And if so, what is the sin and cosine of \(\displaystyle \frac{2\pi}{3}\) ? . . . You should know this
Click to expand...

I would use: \(\displaystyle \L\,\sin\left(\frac{5\pi}{12}\right)\:=\:\sin\left(\frac{3\pi}{12}\,+\,\frac{2\pi}{12}\right)\:=\:\sin\left(\frac{\pi}{4}\,+\,\frac{\pi}{6}\right)\)

Then we have: \(\displaystyle \L\:\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right)\,+\,\sin\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{4}\right) \;= \;\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}\,+\,\frac{1}{2}\cdot\frac{\sqrt{2}}{2} \;=\;\frac{\sqrt{6}\,+\,\sqrt{2}}{4}\)

 
Yeah! I feel pretty silly now. Where was my head? Thank you for your help.
 
Hello, hiroya!

Can someone explain how \(\displaystyle \sin\left(\frac{\pi}{12}\right)\) became \(\displaystyle \frac{\sqrt{6}}{4}\,-\,\frac{\sqrt{2}}{4}\)?
Click to expand...

We have: \(\displaystyle \L\,\frac{\pi}{12}\:=\:\frac{\pi}{3}\,-\,\frac{\pi}{4}\)

Then: \(\displaystyle \L\,\sin\left(\frac{\pi}{12}\right) \:=\:\sin\left(\frac{\pi}{3}\,-\,\frac{\pi}{4}\right)\)

. . \(\displaystyle \L=\;\sin\frac{\pi}{3}\cdot\cos\frac{\pi}{4}\,-\,\sin\frac{\pi}{4}\cdot\cos\frac{\pi}{3}\)

. . \(\displaystyle \L=\;\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}\,-\,\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\)

. . \(\displaystyle \L=\;\frac{\sqrt{6}}{4}\,-\,\frac{\sqrt{2}}{4}\)

 
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