Question

求解：高一二倍角的正弦余弦正切公式填空题

如图求值

Answer 1

（sin50°(1+√3*tan10°)-cos20°）/（cos80°*√（1-cos20°）)
=（sin50°(1+√3*sin10°/cos10°)-cos20°）/（cos80°*√（1-cos20°））
=(sin50°((cos10°+√3*sin10°)/cos10°)-cos20°）/（cos80°*√(2(sin10°)^2))
=(sin50°((1/2)cos10°+(√3/2)*sin10°)/((1/2)*cos10°)-cos20°）/（cos80°*√(2(sin10°)^2))
=(sin50°(sin30°*cos10°+cos30°*sin10°)/(1/2)*cos10°-cos20°）/（(√2)cos80°*sin10°)
=(sin50°*sin40°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=(sin50°*cos50°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=((1/2)sin100°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=((1/2)sin80°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=((1/2)cos10/(1/2)*cos10°-cos20°）/（(√2)sin10°sin10°)
=(1-cos20°)/（(√2)sin10°sin10°)
=2(sin10°)^2/(√2)(sin10°)^2
=2/√2
=√2