Question

Hi! The question I need help with is 6cos^2x-3 and the directions say to use a double-angle formula to write the expression. This is what I have but I don't know if it's right?

6cos^2(x)-3
=3(2cos^2(x)-1)
=3cos(2x)

Answer 1

you are right on.

Answer 2

Okay, thank so much! I have another question: This one says to use the Half-Angle formulas to determine the exact value of sin(pi/12). Here's what I have:

π/12 = ( 180° ) / 12 = 15°.

= sin ( π/12 )

= sin 15°

= sin ( 45° - 30°)

= sin 45°· cos 30° - cos 45°· sin 30°

= (1/√2)·(√3 /2 ) - (1/√2)·(1/2)

= ( √3 - 1 ) / (2√2)