Question

简单的三角恒等变换例题有哪些？

Answer 1

一、解：a·b=λcosα·2sinα+1

=2λsinαcosα+1

=λsin(2α)+1

当2α=π/2+2kπ时

即α=π/4+kπ

a·bmax=λ+1=5

λ=4

二、sinx/cosx=tanx=2/3

sinx=2/3*cosx

代入恒等式sin²x+cos²x=1

(4/9+1)*cos²x=1

cos²x=9/13

2cos²x-1=5/13

扩展资料：

sin2α = 2cosαsinα

= sin²(α+π/4)-cos²(α+π/4)

= 2sin²(a+π/4)-1

= 1-2cos²(α+π/4)

cos2α = cos²α-sin²α

= 1-2sin²α

= 2cos²α-1

= 2sin(α+π/4)·cos(α+π/4)

tan2α = 2tanα/[1-(tanα)²]

参考资料来源：百度百科-三角恒等变形