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# NEED HELP ASAP!!!!!

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cos(pi/6-x) can be expressed in the form A(sin(x))+B(cos(x)) where A and B are constants. Find B/A.

Dec 18, 2020

#1
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By the angle sum/difference formula for cos,

cos(a - b)  =  cos a cos b + sin a sin b

And so...

$${\cos(\frac{\pi}{6}-x)}\ =\ \cos(\frac{\pi}{6})\cos(x)+\sin(\frac{\pi}{6})\sin(x)\\~\\ \phantom{\cos(\frac{\pi}{6}-x)}\ =\ \sin(\frac{\pi}{6})\sin(x)+\cos(\frac{\pi}{6})\cos(x)$$

Now we can see that

$$A\ =\ \sin(\frac{\pi}{6})\ =\ \frac12 \\~\\ B\ =\ \cos(\frac{\pi}{6})\ =\ \frac{\sqrt3}{2}$$

And so...

$$\dfrac{B}{A}\ =\ \dfrac{(\frac{\sqrt3}{2})}{(\frac{1}{2})}\ =\ \sqrt3$$    _

Dec 18, 2020