Question

已知3π/4<α<π,tanα+1/tanα=-10/3.求5sin^2a/2+8sina/2cosa/2+11cos^2a/2-8/

Answer 1

3π/4<a<π
tan3π/4<tana<tanπ
-1<tana<0

tana+1/tana=-10/3
3tan�0�5a+10tana+3=0
(3tana+1)(tana+3)=0
-1<tana<0
tana=-1/3

[5sin�0�5α/2+(8sinα/2)cosα/2+11cos�0�5α/2-8]÷(√2)sin(α-π/2)

=[5+4sinα+6cos�0�5α/2-8]÷(√2)sin(α-π/2)

=[4sinα+6cos�0�5α/2-3]÷(-√2)cosα

=[4sinα+3cosα]÷(-√2)cosα

=-4sinα/√2cosα-3cosα/√2cosα

=-4√2sinα/2cosα-3√2cosα/2cosα

=-2√2sinα/cosα-3√2/2

=-2√2tana-3√2/2

=-2√2*(-1/3)-3√2/2

=3√2/2-3√2/2

=0