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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
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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

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