Question

已知(tanα-3)(sinα+cosα +3)=0,求下列各式的值 4sinα-2sinα/5cosα+3sinα 2

Answer 1

(tanα-3)(sinα+cosα+3)=0

sinα+cosα+30,tanα-3=0,tanα=3.

(4cosα-2sinα)/(5cosα+3sinα)=(4-2tanα)/(5+3tanα)=(4-6)/(5+9)=-1/7

(2/3)(sinα)^2+(1/4)(cosα)^2

=[(2/3)(sinα)^2+(1/4)(cosα)^2]/[(sinα)^2+(cosα)^2]

=[(2/3)(tanα)^2+1/4]/[(tanα)^2+1]

=5/8