(1+tanα)/(1-tanα)=3+2√2,求cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α-π)的值

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(1+tanα)/(1-tanα)=3+2√2,求cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α-π)的值

cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α-π)
=cos^2α +sinαcosα+2sin^2α
=1+sinαcosα+sin^2α
=1+0.5sin2α+sin^2α
=1+tanα/(1+tan^2α)+tan^2α/(1+tan^2α)
=(1+tanα+2tan^2α)/(1+tan^2α)
因为(1+tanα)/(1-tanα)=3+2√2
所以tanα=(√2)/2
那么原式化=(1+√2/2 +2*1/2)/(1+1/2)=(4+√2)/3
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