[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
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[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+3
[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
[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
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，
[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
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[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
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