(1+sin2α) ÷ (2cos²α+sin2α)=0.5tanα+0.5(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)=tan² × tan²

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/16 09:56:51
(1+sin2α)÷(2cos²α+sin2α)=0.5tanα+0.5(3-4cos2A+cos4A)÷(3+4cos2A+cos4A)=tan²×tan²(1+sin

(1+sin2α) ÷ (2cos²α+sin2α)=0.5tanα+0.5(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)=tan² × tan²
(1+sin2α) ÷ (2cos²α+sin2α)=0.5tanα+0.5
(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)=tan² × tan²

(1+sin2α) ÷ (2cos²α+sin2α)=0.5tanα+0.5(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)=tan² × tan²
(1+sin2α) ÷ (2cos²α+sin2α)
=(sin²α+cos²α+2sinαcosα)÷(2cos²α+2sinαcosα)
=(sinα+cosα)²÷[2cosα(cosα+sinα)]
=(sinα+cosα)÷(2cosα)
=0.5sinα/cosα+0.5
=0.5tanα+0.5
(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)
=(3-4cos2A+2cos²2A-1)÷(3+4cos2A+2cos²2A-1)
=(2cos²2A-4cos2A+2)÷(2cos²2A+4cos2A+2)
=(cos²2A-2cos2A+1)÷(cos²2A+2cos2A+1)
=(cos2A-1)²÷(cos2A+1)²
=(1-2sin²A-1)²÷(2cos²A-1+1)²
=(-2sin²A)²÷(2cos²A)²
=(sin²A)²÷(cos²A)²
=tan²A×tan²A