compound formula question

[chhoon]

use an appropriate compound formula to determine an extra value for

sin(31pi/12)

Please help me solve this ! test tomorrow!

 

[mmm4444bot]

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?

 

[soroban]

Hello, chhoon!

$\displaystyle \text{Use an appropriate compound formula to determine the exact value for: }\,\sin\left(\frac{31\pi}{12}\right)$

$\displaystyle \text{Note that: }\:\frac{31}{12} \;=\;2 + \frac{7}{12} \;=\;2 + \frac{4}{12} + \frac{3}{12} \;=\;2 + \frac{1}{3} + \frac{1}{4}$


$\displaystyle \text{So we have: }\;\sin\left(\frac{31\pi}{12}\right) \;=\;\sin\left(2\pi + \frac{\pi}{3} + \frac{\pi}{4}\right) \;=\;\sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right)$


$\displaystyle \text{Now apply: }\;\sin(A+B) \:=\:\sin A\cos B + \cos A\sin B$


$\displaystyle \text{We have: }\;\sin\left(\frac{\pi}{3} + \frac{\pi}{4}\right) \;=\;\sin\frac{\pi}{3}\cos\frac{\pi}{4} + \cos\frac{\pi}{3}\sin\frac{\pi}{4}$

. . . . . . . . . . . . . . . . .$\displaystyle =\;\;\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \;\;+\;\; \frac{1}{2}\cdot\frac{\sqrt{2}}{2}$

. . . . . . . . . . . . . . . . .$\displaystyle =\qquad \frac{\sqrt{6} + \sqrt{2}}{4}$