You began the problem correctly. Continue to use the cos "double angle" formula, applying it twice.
cos4x = cos[2(2x)] = 2 cos^2(2x) - 1
= 2(2cos^2 x - 1)^2 -1
= 2*[4 cos^4 x -4cos^2 x +1) -1
= 8cos^2 x -8cos^2 x +1
Math - Trig - Double Angles - Prove: cos4x = 8cos^4x - 8cos^2x + 1 My Attempt: ...
Trigonometry - Need help solving problems using the Trigonometric Identities: ...
trig - sin^4 x = (3-4cos 2x + cos4x)/8 prove on one side please!
Pre-Calculus - Find all solutions to the equation in the interval [0, 2pi) cos4x...
math - Why does 1/8cos(2y+6) become 1/8cos(arcsinx/4)?
trig help again i cant get any of these - sin^4cos^2x=(1/16)(1-cos 2x)(1-cos4x) ...
trig!! - sin^4 x (cos^2 x)=(1/16)(1-cos 2x)(1-cos4x) work one 1 side only!
Math - Can I please get some help on these questions: 1. How many solutions does...
trig - tan^2(5x)cos^4(5x) = 1/8 - 1/8cos(20x)
trig - tan^2(5x)cos^4(tx = 1/8 - 1/8cos(20x)