!!化简高手来!![cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=
来源:百度知道 编辑:UC知道 时间:2024/04/28 19:57:36
不要这种复制答案
cos40°+sin50°×(1+√3tan10°)
=cos40°+sin50°×(tan60°-tan10°)/tan50°
=cos40°+(tan60°-tan10°)cos50°
=cos40°+√3cos50°-tan10°cos50°
=cos40°+√3sin40°-tan10°sin40°
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40°
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°
=2cos20°-2cos20°+2(cos20°) ∧2
=1+cos40°
sin70°√1+cos40°=sin70°(√2)cos20°=(√2)cos20°∧2
=√2/2(1+cos40°)
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=√2
cos40°+sin50°×(1+√3tan10°)
=cos40°+sin50°×(tan60°-tan10°)/tan50°
=cos40°+(tan60°-tan10°)cos50°
=cos40°+√3cos50°-tan10°cos50°
=cos40°+√3sin40°-tan10°sin40°
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40°
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°
=2cos20°-2cos20°+2(cos20°) ∧2
=1+cos40°
sin70°√1+cos40°=sin70°(√2)cos20°=(√2)cos20°∧2
=√2/2(1+cos40°)
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=√2
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]
=[cos40°+sin50°×(tan60°-tan10°)/tan50°]/sin70°(√2)cos20°
=[cos40°+(tan60°-tan10°)cos50°]/sin70°(√2)cos20°
=[cos40°+√3cos50°-tan10°cos50°]/ (√2)cos20°^2
=[cos40°+√3sin40°-tan10°sin40°] /√2/2(1+cos40°)
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° /√2/2(1+cos40°)
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°/√2/2(1+cos40°)
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°/√2/2(1+cos40°)
=2cos20°-2cos20°+2(cos20°) ^2/√2/2(1+cos40°)
=1+cos40°/√2/2(1+cos40°)
=√2
救命~~~~~~化简求值(cos20°)^4+(cos40°)^4+(cos80°)^4=
已知0°< α< β<90° ,且sinα.sinβ是方程x^2-(√2cos40°)x+cos^2
求值1/tan20*cos10+√3sin 10*tan70-2cos40
cos40°+cos60°+cos80°+cos160°怎样求值?
已知0<a<B<90度,且sina,sinB是x^2-(根号2*cos40度)x+(cos40度)^2-1/2=0的两实数根,求sin(B-5a)的值
请各位高手帮帮小妹!cos40度(1+根号3乘以cot80度)=
sin^2x+2sinxcosx求周期~~~~如何化简?
sin(pi/2-a)怎么化简
3csc^2 20°-sec^2 20°=32cos40°
SIN(36度)如何求出来的