solve:
cos^4(x)-sin^4(x)=cos(2x)
cos(4x)=cos^2(x)-sin^2(x)
2 answers
pls help my teacher didn't teach us anything
seems like I just did these already
Remember your double-angle formulas
cos(2θ) = cos^2(θ) - sin^2(θ)
so the first one is just an identity
cos^4(x)-sin^4(x)
= (cos^2x + sin^2x)(cos^2x - sin^2x)
= 1 * (cos^2x) - sin^2x)
= cos(2x)
as written, the next one is not an identity
cos(4x) = 2cos^2(2x) - 1
so now you have
2cos^2(2x) - 1 = cos(2x)
2cos^2(2x) - cos(2x) - 1 = 0
(2cos(2x) + 1)(cos(2x - 1) = 0
so,
cos(2x) = -1/2
cos(2x) = 1
2x = 0+2kπ
x = kπ
or
2x = 2π/3 + 2kπ or 4π/3+2kπ
x = π/3 + kπ or 2π/3 + kπ = kπ ± π/3
Remember your double-angle formulas
cos(2θ) = cos^2(θ) - sin^2(θ)
so the first one is just an identity
cos^4(x)-sin^4(x)
= (cos^2x + sin^2x)(cos^2x - sin^2x)
= 1 * (cos^2x) - sin^2x)
= cos(2x)
as written, the next one is not an identity
cos(4x) = 2cos^2(2x) - 1
so now you have
2cos^2(2x) - 1 = cos(2x)
2cos^2(2x) - cos(2x) - 1 = 0
(2cos(2x) + 1)(cos(2x - 1) = 0
so,
cos(2x) = -1/2
cos(2x) = 1
2x = 0+2kπ
x = kπ
or
2x = 2π/3 + 2kπ or 4π/3+2kπ
x = π/3 + kπ or 2π/3 + kπ = kπ ± π/3
1 answer