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)
2 answers
you are right on.
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)
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)
3 answers