证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tanα)
证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tanα)
证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tanα)
证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tanα)
证明:sin(α+3π/4)*cos(α+π/4)=sin[π/2+(α+π/4)]*cos(α+π/4)=cos(α+π/4)*cos(α+π/4)=cos��(α+π/4)=[√2/2*(cosα-sinα)]��=1/2(cosα-sinα)��;
2-2sin(α+3π/4)*cos(α+π/4)=2-(cosα-sinα)��=2-(cos��α+sin��α-2cosαsinα)=1+2cosαsinα=cos��α+sin��α+2cosαsinα=(cosα+sinα)��;
cos^4α-sin^4α=(cos��α)��-(sin��α)��=(cos��α-sin��α)(cos��α+sin��α)=(cosα-sinα)(cosα+sinα);
[2-2sin(α+3π/4)*cos(α+π/4)]/(cos^4α-sin^4α)=(cosα+sinα)��/(cosα-sinα)(cosα+sinα)=(cosα+sinα)/(cosα-sinα)=(1+tanα)/(1-tanα)
[2-2sin(a+3π/4)cos(a+π/4)]/(cos^4a-sin^4a)
=[2-2(-√2/2*sina+√2/2*cosa)(√2/2*cosa-√2/2*sina)]/[(cos²a+sin²a)(cos²a-sin²a)]
=[2-(cosa-sina)²]/(cos²a-sin²a)
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[2-2sin(a+3π/4)cos(a+π/4)]/(cos^4a-sin^4a)
=[2-2(-√2/2*sina+√2/2*cosa)(√2/2*cosa-√2/2*sina)]/[(cos²a+sin²a)(cos²a-sin²a)]
=[2-(cosa-sina)²]/(cos²a-sin²a)
=(1+sin2a)/cos2a;用万能公式
=[1+2tana/(1+tan²a)]/[(1-tan²a)/(1+tan²a)]
=(1+tana)²/(1-tan²a)
=(1+tana)/(1-tana);得证!
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