Expert: Sherry Wallin - 4/2/2011

Question
QUESTION: 1)Using an appropriate double-angle formula ,expres cos(4theta)in terms of cos (2theta). Using your answer and another appropriate double-angle formula, derive the formula cos(4theta)=2cos^4theta-4cos^2(theta)sin^2(theta)+2sin^4theta-1

2) given that 1/2pi<theta<pi and cos(theta)=-1/5 , use an appropriate trigonometric formulas to find the exact values of the following : cos(2theta),sin(theta) and sin(2theta)

ANSWER: Elle~

First let's use x for theta so we have

cos(4theta) = cos(4x)
= cos[2(2x)]
= 2cos^2(2x) - 1
= 2cos(2x)*cos(2x) - 1
= 2[(cos^2(x) - sin^2(x))(cos^2(x) - sin^2(x))] - 1
= 2[cos^4(x) - 2cos^2(x)sin^2(x) + sin^4(x)] - 1
= 2cos^4(x) - 4cos^2(x)sin^2(x) + 2sin^4(x) - 1

which is what you wanted to show, right?


Math Prof

---------- FOLLOW-UP ----------

QUESTION: Using an appropriate double-angle formula ,expres cos(4theta)in terms of cos (2theta).

Answer
Elle~

I thought this was answered but I guess it is still not clear. What I am going to use as an identity is cos(2x) = 2cos^2(x) - 1:
cos(4x) = cos(2*2x) = cos(2y) where y = 2x thus cos(2y) = 2cos^2(y) - 1 by the identity above but now replace the y with 2x getting
cos(4x) = cos(2*2x) = cos(2y) = 2cos^2(y) - 1 = 2cos^2(2x) - 1

Is this clearer?

Math Prof