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.
+9466 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...
\(\dfrac{B}{A}\ =\ \dfrac{(\frac{\sqrt3}{2})}{(\frac{1}{2})}\ =\ \sqrt3\) _
