已知sin^4θ+cos^4θ=5/9,则sin2θ由sin^4θ+cos^4θ=,∴(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9.∴2sin^2θcos^2θ=.∴sin2^2θ=8/9.(不是说2sinθcosθ=sin2θ吗,那这里为什么是这样?)

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已知sin^4θ+cos^4θ=5/9,则sin2θ由sin^4θ+cos^4θ=,∴(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9.∴2sin^2θcos^2θ=.∴sin2

已知sin^4θ+cos^4θ=5/9,则sin2θ由sin^4θ+cos^4θ=,∴(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9.∴2sin^2θcos^2θ=.∴sin2^2θ=8/9.(不是说2sinθcosθ=sin2θ吗,那这里为什么是这样?)
已知sin^4θ+cos^4θ=5/9,则sin2θ
由sin^4θ+cos^4θ=,
∴(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9.
∴2sin^2θcos^2θ=.∴sin2^2θ=8/9.(不是说2sinθcosθ=sin2θ吗,那这里为什么是这样?)

已知sin^4θ+cos^4θ=5/9,则sin2θ由sin^4θ+cos^4θ=,∴(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9.∴2sin^2θcos^2θ=.∴sin2^2θ=8/9.(不是说2sinθcosθ=sin2θ吗,那这里为什么是这样?)
那就这样
2sin^2(θ)·cos^2θ=2(sinθ·cosθ)^2
∵2sinθ·cosθ=sin2θ
∴2sin^2(θ)·cos^2(θ)=2sin^2(2θ)

问老师

先平方,再利用和差公式

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