Question

已知cos(4分之π+α)=5分之3,12分之17π<α<4分之7π,求1-tanα分之sin2α(1+tanα)的值

Answer 1

解：
cos(π/4+α)=3/5
cos(π/4)cosα-sin(π/4)sinα=3/5
[(√2)/2](cosα-sinα)=3/5
cosα-sinα=(3√2)/5………………………………(1)
(cosα-sinα)²=[(3√2)/5]²
2sinαcosα=7/25…………………………………(2)
1+2sinαcosα=32/25
sin²α+2sinαcosα+cos²α=32/25
(cosα+sinα)²=32/25
cosα+sinα=±(4√2)/5……………………………(3)

sin2α(1+tanα)/(1-tanα)
=2sinαcosα(cosα+sinα)[±(4√2)/5]/(cosα-sinα)
=(7/25))[±(4√2)/5]/[(3√2)/5]
=(7/25))[±(4√2)/5]×[5(√2)/6]
=(7/25))(±4/3)
=±28/75