Question

tanα=1/2,求①sin^2 α+2cos^2 α ②sinα*cosα+2cos^2α

Answer 1

解

sin²a+2cos²a

=(sin²a+2cos²a)/(sin²a+cos²a)——除以sin²a+cos²a=1,值不变

=(tan²a+2)/(tan²a+1)——分子分母同时除以cos²a

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

=9/4×4/5

=9/5

sinacosa+2cos²a

=(sinacosa+2cos²a)/(sin²a+cos²a)——与第一题同理

=(tana+2)/(tan²a+1)

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

=5/2×4/5

=2