(1+sin2α) ÷ (2cos²α+sin2α)=0.5tanα+0.5(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)=tan² × tan²
问题描述:
(1+sin2α) ÷ (2cos²α+sin2α)=0.5tanα+0.5
(3 - 4cos2A + cos4A)÷ (3 + 4cos2A + cos4A)=tan² × tan²
答
(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...