Question
please help me with this proving question
prove that
4cos(theta) * cos(theta + 120) * cos(theta - 120) = cos(3theta)
PLEASE HELP THANKS
prove that
4cos(theta) * cos(theta + 120) * cos(theta - 120) = cos(3theta)
PLEASE HELP THANKS
Answers
drwls
Use the cos (A+B) and cos (A-B) formulas to rewrite the left hand side in terms of sin theta and cost theta.
The product (cos(A+B)*cos(A-B) can be written
(cos A cos B)^2 - (sinA sinB)^2
which when B = 120 is
(1/4)cos^2A -(3/4)sin^2A
= (1/4)cos^2A -(3/4)(1 - cos^2A)
= cos^2A - (3/4)
So the left side becomes
4 cos^3(theta)-3 cos theta
Now you work on the right hand side, treating cos (3 theta) as cos (theta + 2 theta) and provde it equals the left hand side.
The product (cos(A+B)*cos(A-B) can be written
(cos A cos B)^2 - (sinA sinB)^2
which when B = 120 is
(1/4)cos^2A -(3/4)sin^2A
= (1/4)cos^2A -(3/4)(1 - cos^2A)
= cos^2A - (3/4)
So the left side becomes
4 cos^3(theta)-3 cos theta
Now you work on the right hand side, treating cos (3 theta) as cos (theta + 2 theta) and provde it equals the left hand side.
Vincent
damn thanks
drwls
I converted the more difficult half of the equation for you. The answer is correct. You should have very little trouble proving the right side is the same.