trig
how can you confirm the identity
cos^4x = (1/8)(3+ 4cos2x+ cos4x)
and
sin4x = (4sinxcosx)(2cos^2x1)

You use the identity:
exp(ix) = cos(x) + i sin(x)
You can then write:
cos(x) = [exp(ix)+ exp(ix)]/2
Take the fourth power:
cos^4(x) =
1/16 [exp(4 i x) + 4 exp(2 i x) + 6 +
4 exp(2 i x) + exp(4 i x)] =
1/8 {3 + [exp(4 i x) + exp(4 i x)]/2 + 4 [exp(2 i x) + exp(2 i x)]/2} =
1/8 [3 + cos(4 x) + 4 cos(2 x)]posted by Count Iblis

sin(4x) can be expanded in the usual way by using the doubling formulas:
sin(4x) = 2 sin(2x) cos(2x) =
4 sin(x) cos(x)cos(2x)=
4 sin(x) cos(x) [2 cos^2(x)  1]
The general method also works, but it slightly more laborious in this simple case:
You start with
exp(ix) = cos(x) + i sin(x)
and take the fourth power of both sides and take the imaginary part:
Im[exp(4 i x)] = 4 cos^3(x)sin(x) 
4 cos(x) sin^3(x)
Im[exp(4 i x)] = sin(4 x), so we have:
sin(4 x) =
4 cos(x)sin(x)[cos^2(x) sin^2(x)]
Using that sin^2(x) = 1  cos^2(x) gives you the desired result.posted by Count Iblis
Respond to this Question
Similar Questions

Math  Trig  Double Angles
Prove: cos4x = 8cos^4x  8cos^2x + 1 My Attempt: RS: = 4cos^2x (2cos^2x  1) + 1 = 4 cos^2x (cos2x) + 1 LS: = cos2(2x) = 2cos^2(2x)  1 = (cos^2(2))  cos^2(2x))  1  Prove: 8cos^4x = cos4x + 4cos2x + 3 My Attempt: RS: = 
precalc
use power reducing identities to prove the identity sin^4x=1/8(34cos2x+cos4x) cos^3x=(1/2cosx) (1+cos2x) thanks :) 
Trig identity.
I need help with verifying these trig identities: 1) sin4x = 4sinxcos  8sin^3 x cos x 2) cos3x = cos^3 x  3sin^2 x cos x 
Calculus
Prove the identity and give its domain. 4sinxcosx+sin4x=8sinxcos^3x 
Math  Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x)  1 cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1 cos^2(x)(1  
Trig
cos4x*cos3x + sin4x*sin3x 
Trig.
Prove sin(4x)= (4sinxcosx)(2cos(x)^(2)1 
MATH
veify that each equation is an identity cos2x+tan2xcos2x=1 sin4xcos4x/sin2x=1cot2x 
Calculus
Evaluate the integral e^x sin4x dx. I know the answer is 1/17 e^x sin4x  4/17 e^x cos4x + C but I don't know how to solve it out. 
Math (Trig)
Rewrite the following expression in terms of tan3x: (sin2x + sin4x)/(cos2x + cos4x)