1. ## Trig value question

I would like to know if I am doing these correctly. I am supposed to find exact solutions of pi/6,pi/4 for these problems.
cos(11pi/12), tan(pi/12), sin(5pi/12) This is one of the ones I did.

Cos(11pi/12)= Cos(3pi/4 + pi/6) - Sin(3pi/4 + pi/6)
Cos(3pi/4) Cos(pi/6) - Sin(3pi/4) Sin(pi/6)
-sqrt(2)/2 *sqrt(3)/2 - Sqrt(2)/2 *1/2 = -sqrt(6)/2 -sqrt(2)/2
Also if this is correct, do I use cotangent with tan.

Thank you,
Keith Stevens

2. Originally Posted by kcsteven
I would like to know if I am doing these correctly. I am supposed to find exact solutions of pi/6,pi/4 for these problems.
cos(11pi/12), tan(pi/12), sin(5pi/12) This is one of the ones I did.

Cos(11pi/12)= Cos(3pi/4 + pi/6) - Sin(3pi/4 + pi/6)
Cos(3pi/4) Cos(pi/6) - Sin(3pi/4) Sin(pi/6)
-sqrt(2)/2 *sqrt(3)/2 - Sqrt(2)/2 *1/2 = -sqrt(6)/2 -sqrt(2)/2
Also if this is correct, do I use cotangent with tan.

Thank you,
Keith Stevens
I'm curious, where did you get that first line from?

$cos \left ( \frac{3\pi}{4} + \frac{\pi}{6} \right ) = cos \left ( \frac{3\pi}{4} \right ) cos \left ( \frac{\pi}{6} \right ) - sin \left ( \frac{3\pi}{4} \right ) sin \left ( \frac{\pi}{6} \right )$
$sin \left ( \frac{3\pi}{4} + \frac{\pi}{6} \right ) = sin \left ( \frac{3\pi}{4} \right ) cos \left ( \frac{\pi}{6} \right ) + sin \left ( \frac{\pi}{6} \right ) cos \left ( \frac{3\pi}{4} \right )$
I would use $tan \left ( \frac{\pi}{12} \right ) = tan \left ( \frac{1}{2} \cdot \frac{\pi}{6} \right )$