cos10t(cos10^4t)^2的傅里叶变换
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/25 11:16:09
cos10/tan20+√3sin10tan70-2cos40=cos10/cot70+√3sin10tan70-2cos40=cos10tan70+√3sin10tan70-2cos40=tan70
解sin10°-根号3*cos10°=sin10°-√3*cos10°=2(1/2sin10°-√3/2*cos10°)=2(cos60°sin10°-sin60°*cos10°)=2cos(60°+
sin20=2sin10cos10cos20=1-2(sin10)^2原式=2(sin10)^2/[1-2(sin10)^2]=-1+1/[1-2(sin10)^2]=-1+1/cos20=sec20
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(sin10/cos10*cos10-根号3*cos10)/sin50=(sin10-根号3*cos10)/sin50=2(1/2*sin10-根号3/2*cos10)/sin50=2(cos60*s
应该是tan70°*cos10°*(√3*tan20°-1)吧=tan70°*cos10°*(tan60°*tan20°-1)=tan70°*cos10°*[(sin60°*sin20°/cos60°
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cos10°=2cos^2(5°)-11+cos10°=2cos^2(5°)√1+cos10°=√2cos5°
(1)(2cos10°-sin20°)/sin70°=[2cos(30°-20°)-sin20°]/sin(90°-20°)=(2cos30°cos20°+2sin30°sin20°-sin20°)/
1.ctg10-4cos10=(cos10-4sin10cos10)/sin10=(cos10-2sin20)/sin10=(sin80-sin20-sin20)/sin10=(2cos50sin30
cos10°tan20°+3sin10°•tan70°-2cos40°=cos10°tan20°+3sin10°tan20°-2cos40°=2sin(10°+30°)tan20°-2cos40°=2
f(80)-4cos10=8sin80-tan80-4cos10=8cos10-cot10-4cos10=4cos10-cos10/sin10=(4sin10cos10-cos10)/sin10=(2
(tan10°-√3)×cos10°/sin50°=(sin10°/cos10°-√3)×cos10°/sin50°=(sin10°-√3cos10°)/sin50°=2(cos60°sin10°-s
cos20=cos(2*10)=(cos10)^2-(sin10)^2所以cos20/(cos10+sin10)+sin10=[(cos10)^2-(sin10)^2]/(cos10+sin10)+s
因为2cos10*sin10=sin20,即cos10*sin10=1/2sin20代入左边即可
按题意理解应该是这样:|sin10度-cos10度|/(sin10度-cos10度)=-1∵sin10º
sin25π/6+cos10π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)=sin(4π+π/6)+cos(4π-2π/3)+tan(-6π-π/4)+sin(-2π-π
(本题满分8分)原式=cos20°cos10°sin20°+3sin10°sin70°cos70°−2cos40°----------------------(1分)=cos20°cos10°+3si