已知sin^4θ+cos^4θ=5/9,则sin2θ=?

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已知sin^4θ+cos^4θ=5/9,则sin2θ=?已知sin^4θ+cos^4θ=5/9,则sin2θ=?已知sin^4θ+cos^4θ=5/9,则sin2θ=?sin^4θ+cos^4θ=[(

已知sin^4θ+cos^4θ=5/9,则sin2θ=?
已知sin^4θ+cos^4θ=5/9,则sin2θ=?

已知sin^4θ+cos^4θ=5/9,则sin2θ=?
sin^4θ+cos^4θ
=[(sinθ)^2+(cosθ)^2]^2-2[(sinθ)^2][(cosθ)^2]
=1-2[(sinθ)^2][(cosθ)^2]
=1-1/2(sin2θ)^2
=5/9
所以(sin2θ)^2=8/9
sin2θ=2√2/3或sin2θ= -2√2/3

sin^4θ+cos^4θ
=[(sinθ)^2+(cosθ)^2]^2-2[(sinθ)^2][(cosθ)^2]
=1-1/2(sin2θ)^2
=5/9
(sin2θ)^2=8/9
sin2θ=2√2/3、sin2θ= -2√2/3