Question

[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=？

求过程和答案，哪位大神帮帮忙哈，谢谢啦

Answer 1

[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]
=[cos(90°-(45°+θ))]/[cos(45°-θ)]-[sin(90°-(θ+60°))]/[sin(θ+30°)]
=cos(45°-θ)/cos(45°-θ)-sin(30°-θ)/sin(θ+30°)

=1-sin(30°-θ)/sin(θ+30°)
题有没有错

Answer 2

这个问题你到三角函数的积化和差、积差化积的公式上面看看，应该就可以弄出来了，

Answer 3

[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]
=[sin(45°+θ)]/[sin(45°+θ)]-[sin(θ+30°)]/[sin(θ+30°)]
=1-1
=0