早教吧作业答案频道 -->数学-->
化简:cot(θ+4π)*cos(θ+π)*sin(θ+3π)/tan(π+θ)*cos^3(-π-θ)求值:√1-2sin10°cos10°/cos10°-√1-cos^2170°求证:1-2sin2x*cos2x/cos^22x-sin^22x=1-tan2x/1+tan2x
题目详情
化简:cot(θ+4π)*cos(θ+π)*sin(θ+3π)/tan(π+θ)*cos^3(-π-θ)
求值:√1-2sin10°cos10°/cos10°-√1-cos^2 170°
求证:1-2sin2x*cos2x/cos^2 2x-sin^2 2x=1-tan2x/1+tan2x
求值:√1-2sin10°cos10°/cos10°-√1-cos^2 170°
求证:1-2sin2x*cos2x/cos^2 2x-sin^2 2x=1-tan2x/1+tan2x
▼优质解答
答案和解析
cot(θ+4π)*cos(θ+π)*sin(θ+3π)/tan(π+θ)*cos^3(-π-θ)=cotθ(-cosθ)(-sinθ)/tanθcos^3θ=cos^2θ/sinθcos^2θ=1/sinθ=cscθ
√1-2sin10°cos10°/cos10°-√1-cos^2 170°=√sin^2 10°+cos^2 10°-2sin10°cos10°/cos10°-√sin^2 170°+cos^2 170°-cos^2 170°=√(sin10°-cos10°)^2/cos10°-√sin^2 170°=cos10°-sin10°/cos10°-sin170°=cos10°-sin10°/cos10°-sin10°=1
1-2sin2x*cos2x/cos^2 2x-sin^2 2x=cos^2 2x-2sin2xcos2x+sin^2 2x/cos^2 2x-sin^2 2x=1-2tan2x+tan^2 2x/1-tan^2 2x=(1-tan2x)^2/(1-tan2x)(1+tan2x)=1-tan2x/1+tan2x
√1-2sin10°cos10°/cos10°-√1-cos^2 170°=√sin^2 10°+cos^2 10°-2sin10°cos10°/cos10°-√sin^2 170°+cos^2 170°-cos^2 170°=√(sin10°-cos10°)^2/cos10°-√sin^2 170°=cos10°-sin10°/cos10°-sin170°=cos10°-sin10°/cos10°-sin10°=1
1-2sin2x*cos2x/cos^2 2x-sin^2 2x=cos^2 2x-2sin2xcos2x+sin^2 2x/cos^2 2x-sin^2 2x=1-2tan2x+tan^2 2x/1-tan^2 2x=(1-tan2x)^2/(1-tan2x)(1+tan2x)=1-tan2x/1+tan2x
看了 化简:cot(θ+4π)*c...的网友还看了以下:
设集合s={0 1 2 3 4 5} A是s的一个子集当x属於A 时 若有x-1不属於A且x+1不 2020-04-06 …
s=0:t=o:u=0fori=1to3forj=1toifork=jto3s=s+1nextkt 2020-05-14 …
一直概率密度,求分布函数的一道题目图中画红框的那个,当x>=1时,f(x)不是等于0吗?为什么还有 2020-05-15 …
GUI中分段函数如何绘制本人matlab新手,想用gui做一个软件,但用axes功能的时候,分段函 2020-05-17 …
冰期相对水深0.0,0.2,0.4,0.6,0.8,1.0的测点流速分别为0.30m/s,0.42m 2020-05-27 …
高二数学,大神进!已知椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的离心率为√2/2, 2020-06-27 …
设S={0,1,1/2,.1/N.},H是S的一个开覆盖,证明:H中存在有限个开区间覆盖S 2020-07-04 …
3.对单底物酶按Michaelis公式,当V分别为0.9Vmax和0.1max时,它们相应的底物浓 2020-07-20 …
集合s={0,1,2,3,4,5},A是S的一个子集,当x∈A时,若有x-1不属于A且x+1不属于 2020-07-29 …
1.集合S={0,1,2,3,4,5},A是S的一个子集,当x∈A时,若有x-1不属于A且x+1不 2020-07-30 …