Question

已知tan2α=-sinα,α∈（pai/2,pai),求tanα的值

Answer 1

tan2a=-sina

sin2a/cos2a=-sina

2cosa/cos2a=-1

cos2a/cosa=-2

2cosa-1/cosa=-2

2cosa^2-1+2cosa=0

(cosa-1)^2=1/2+1/4

cosa=1-√3/2 cosa^2=7/4-√3 sina^2=1-(7/4-√3)=√3-3/4

tana^2=sina^2/cosa^2=(4√3-3)/(7-4√3)=(4√3-3)(7+4√3)=27+16√3

tana=-√(27+6√3)