Question

数学解三角函数

已知tanα/（tanα-1）=-1，求下列各式的值：
（2）sin^2 α+sin αcosα+2.

Answer 1

tanα/（tanα-1）=-1
=>tanα=1/2
原式除个cosα^2
sin^2 α+sin αcosα+2=cosα^2（tanα^2+tanα）+2=3/4cosα^2+2
cosα^2=1/(tanα^2+1)=4/5
所以原式=3/4*4/5+2=13/5

Answer 2

解：

tanα/(tanα-1)=-1

∴tanα=1/2

sin²α+sinαcosα+2

=（sin²α+sinαcosα+2sin²α+2cos²α)/(sin²α+cos²α)

=(3sin²α+2cos²α+sinαcosα)/(sin²α+cos²α)

=(3tan²α+2+tanα)/(tan²α+1)

=(3/4+2+1/2)/(5/4)

=13/5