Question

三角恒等变形求解

Answer 1

sin50°*(1+根号3×tan10°)
=sin50°*(1+根号3*sin10°/cos10°)
=sin50°/cos10°*(cos10°+根号3*sin10°)
=2*sin50°/cos10°*(1/2*cos10°+根号 3/2*sin10°)
=2*sin50°/cos10°*(sin30°*cos10°+cos30°*sin10°)
=(2*sin50°/cos10°)* sin40°
=(2*cos40°/cos10°)* sin40°
=(2* sin40°cos40°)/ cos10°
=sin80°/cos10°
=cos10°/cos10°
=1

Answer 2

sin50°（1+根号3 tan10°）
=sin50*(根号3*sin10+cos10)/cos10
=sin50/cos10*2*(cos30sin10+cos10sin30)
=sin50/cos10*2*sin40
=2*sin50*cos50/cos10
=sin100/cos10
=cos10/cos10
=1