compound formula question
- Thread starter chhoon
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mmm4444bot
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I'm not sure what an "extra value" is, but there are different ways to use compound-angle formulas to evaluate sin(31Pi/12).
EG:
31Pi/12 = 40Pi/12 - 9Pi/12
I mean, you could use the formula for sin(u - v) here, yes?
Hello, chhoon!
\(\displaystyle \text{Note that: }\:\frac{31}{12} \;=\;2 + \frac{7}{12} \;=\;2 + \frac{4}{12} + \frac{3}{12} \;=\;2 + \frac{1}{3} + \frac{1}{4}\)
\(\displaystyle \text{So we have: }\;\sin\left(\frac{31\pi}{12}\right) \;=\;\sin\left(2\pi + \frac{\pi}{3} + \frac{\pi}{4}\right) \;=\;\sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right)\)
\(\displaystyle \text{Now apply: }\;\sin(A+B) \:=\:\sin A\cos B + \cos A\sin B\)
\(\displaystyle \text{We have: }\;\sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right) \;=\;\sin\frac{\pi}{3}\cos\frac{\pi}{4} + \cos\frac{\pi}{3}\sin\frac{\pi}{4}\)
. . . . . . . . . . . . . . . . .\(\displaystyle =\;\;\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \;\;+\;\; \frac{1}{2}\cdot\frac{\sqrt{2}}{2}\)
. . . . . . . . . . . . . . . . .\(\displaystyle =\qquad \frac{\sqrt{6} + \sqrt{2}}{4}\)
\(\displaystyle \text{Note that: }\:\frac{31}{12} \;=\;2 + \frac{7}{12} \;=\;2 + \frac{4}{12} + \frac{3}{12} \;=\;2 + \frac{1}{3} + \frac{1}{4}\)
\(\displaystyle \text{So we have: }\;\sin\left(\frac{31\pi}{12}\right) \;=\;\sin\left(2\pi + \frac{\pi}{3} + \frac{\pi}{4}\right) \;=\;\sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right)\)
\(\displaystyle \text{Now apply: }\;\sin(A+B) \:=\:\sin A\cos B + \cos A\sin B\)
\(\displaystyle \text{We have: }\;\sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right) \;=\;\sin\frac{\pi}{3}\cos\frac{\pi}{4} + \cos\frac{\pi}{3}\sin\frac{\pi}{4}\)
. . . . . . . . . . . . . . . . .\(\displaystyle =\;\;\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \;\;+\;\; \frac{1}{2}\cdot\frac{\sqrt{2}}{2}\)
. . . . . . . . . . . . . . . . .\(\displaystyle =\qquad \frac{\sqrt{6} + \sqrt{2}}{4}\)