Question

求三角函数

1. tan19 + tan41 +√3(tan19 * tan41)
2. tan70 + tan50 - √3(tan70 * tan50)
3. sin40(tan10-√3)

数字都表示角度，角度符号没标

Answer 1

1.
解：
∵(tan19°+tan41°)/(1-tan19°*tan41°)=tan(19°+41°)=tan60°=√3
∴tan19°+tan41°=√3(1-tan19°*tan41°)=√3-√3tan19°*tan41°
∴原式=√3-√3tan19°*tan41°+√3tan19°*tan41°=√3
2.
解：
∵(tan70°+tan50°)/(1-tan70°*tan50°)=tan(70°+50°)=tan120°= -√3
∴tan70°+tan50°= -√3(1-tan70°*tan50°)= -√3+√3tan70°*tan50°
∴原式= -√3+√3tan19°*tan41°-√3tan19°*tan41°= -√3
3.
解：原式
=sin40°(sin10°/cos10°-√3cos10°/cos10°)
=(sin40°/cos10°)(sin10°-√3cos10°)
=(sin40°/cos10°)*2sin(10°-60°)
=(sin40°/cos10°)*2sin(-50°)
=-2sin40°sin50°/cos10°
=[(cos40°cos50°-sin40°sin50°)-(cos40°cos50°+sin40°sin50°)]/cos10°
=[cos(40°+50°)-cos(40°-50°)]/cos10°
=[cos90°-cos(-10°)]/cos10°
=(0-cos10°)/cos10°
=-cos10°/cos10°
=-1
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