Question

Please help me

Answer 1

\[\Large \cos\left(\frac{\pi}{6} - x\right) = \cos\left(\frac{\pi}{6}\right)*\cos\left(x\right)-\sin\left(\frac{\pi}{6}\right)*\sin\left(x\right)\]

\[\Large \cos\left(\frac{\pi}{6} - x\right) = \frac{\sqrt{3}}{2}*\cos\left(x\right)-\frac{1}{2}*\sin\left(x\right)\]

\[\Large \cos\left(\frac{\pi}{6} - x\right) = \frac{1}{2}*\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)\]

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Notes:


Step 1: I used the identity cos(X+Y) = cos(X)*cos(Y) - sin(X)*sin(Y)
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf


Step 2: I used the unit circle to evaluate cos(pi/6) and sin(pi/6)
http://www.regentsprep.org/regents/math/algtrig/att5/600px-Unit_circle_angles_svg.jpg


Step 3: This step is optional really but they decided to factor out 1/2, so why not

Answer 2

So, is it the last one?

Answer 3

re-read what I posted

Answer 4

No, what I meant is that that is the answer that I believe is the right answer and I just wanted to check it to see if I was right

Answer 5

The answer isn't choice D