tan(π-θ)=log2¼ 5
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tan( π/4 + θ ) + tan( π/4 - θ ) = 4
[ sin( π/4 + θ )cos( π/4 - θ ) + cos( π/4 + θ )sin( π/4 - θ )] / [cos( π/4 + θ )cos( π/4 - θ )] = 4
sin( π/2 ) / [cos( π/4 + θ )cos( π/4 - θ )] = 4
cos( π/4 + θ )cos( π/4 - θ ) = 1 / 4
( cos π/2 + cos 2θ ) / 2= 1 / 4
cos 2θ = 1 / 2
sin 2θ = √( 1 - cos² 2θ ) = √3 / 2
sin² θ - 2sin θ cos θ - cos² θ = - ( sin 2θ + cos 2θ ) = - (√3 + 1) / 2
已知tanα=-1/2 则(1+2sinα*cosα)/(sin²α-cos²α)的值是
已知tanα=2,求(1+2sinαcosα)/(sin²α-cos²α)的值
已知tanα=4,则2sinα ²+sinαcosα+3=?
已知[(2sin²α+2sinαcosα)÷1+tanα)]=k,试用k表示sinα-cosα的值
已知tanα=2,求sin²α+2sinαcosα+2的
[ sin( π/4 + θ )cos( π/4 - θ ) + cos( π/4 + θ )sin( π/4 - θ )] / [cos( π/4 + θ )cos( π/4 - θ )] = 4
sin( π/2 ) / [cos( π/4 + θ )cos( π/4 - θ )] = 4
cos( π/4 + θ )cos( π/4 - θ ) = 1 / 4
( cos π/2 + cos 2θ ) / 2= 1 / 4
cos 2θ = 1 / 2
sin 2θ = √( 1 - cos² 2θ ) = √3 / 2
sin² θ - 2sin θ cos θ - cos² θ = - ( sin 2θ + cos 2θ ) = - (√3 + 1) / 2
已知tanα=-1/2 则(1+2sinα*cosα)/(sin²α-cos²α)的值是
已知tanα=2,求(1+2sinαcosα)/(sin²α-cos²α)的值
已知tanα=4,则2sinα ²+sinαcosα+3=?
已知[(2sin²α+2sinαcosα)÷1+tanα)]=k,试用k表示sinα-cosα的值
已知tanα=2,求sin²α+2sinαcosα+2的
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