mpjx.net
当前位置:首页 >> sin³π/2等于多少 >>

sin³π/2等于多少

cos(³/₂π+α)=cos[π/2+(π+α)]=公式六=-sin(π+α)=公式二=sinα sin(³/₂π+α)=sin[π/2+(π+α)]=公式六=cos(π+α)=公式二=-cosα

[sin(kx)]'=kcos(kx)=ksin(kx+π/2) [sin(kx)]''=-k²sin(kx)=k²sin(kx+π) [sin(kx)]'''=-k³cos(kx)=k³sin(kx+3π/2) [sin(kx)]^(4)=k^4sin(kx)=k^4sin(kx+2π) ................. [sin(kx)]^(n)=k^nsin(kx+nπ/2) 希望可以帮到...

∫ρ³√(1-ρ²)dρ 令ρ=sinα,dρ=cosαdα 原式=∫sin³αcosαcosαdα =∫sin³...

f(x)=sin²x-sin²(x-π/6) =[sinx-sin(x-π/6)][sinx-sin(x-π/6)] =2sin[(x+x-π/6)/2]cos[(x-x+π/6)/2]·2cos[(x+x-π/6)/2]sin[(x-x+π/6)/2] =sin(2x-π/6)·sinπ/6 =½sin(2x-π/6) ∴T=2π/2=π x∈[-π/3,π/4] 2x-π/6∈[-π/2,2π/3] ∴f(x)最...

解:根据题意分析知,所围成的闭区域在xy平面的投影是圆S:x²+y²=2y 则 ∫∫∫zdxdydz=∫dθ∫rdr∫zdz (作柱面坐标变换) =∫dθ∫[(4r²sin²θ-r^4)/2]rdr =∫dθ∫(2r³sin²θ-r^5/2)dr =∫[8(sinθ)^6-(16/3)(sinθ)^6]dθ =∫[(8/3)...

3、6sin³B-(cos²A)²=6 6sin³B=6+(cos²A)²≥6 则sinB=1,cosA=0 B=2kπ+π/2,A=kπ+π/2(k为任意整数) 4、已知 0lg(1-sin²A)=m-n =>lgcos²A=m-n 因0

解: 因为θ∈[π/4,π/2],sin2θ=3√7 /8 所以cos2θ=-√[1-(sin2θ)²]=-1/8 因为cos2θ=1-2sin²θ=-1/8 所以2sin²θ=9/8 sin²θ=9/16 解得sinθ=3/4或sinθ=-3 /4【舍去】 所以sinθ=3/4 希望可以帮到你 祝学习快乐 O(∩_∩)O~ 打字不易,如...

f'(x)=(sinx-xcosx)/sin²x f''(x)=[(cos-cosx+xsinx)sin²x-2sinxcosx(sinx-xcosx)]/sin⁴x =(xsin²x-2sinxcosx+2xcos²x)/sin³x =(x+xcos²x-sin2x)/sin³x 令g(x)=x+xcos²x-sin2x g'(x)=1+cos²...

解由sin(π-α)-cos(-α)=1/2 得sina-cosa=1/2 平方得1-2sinacosa=1/4 即sinacosa=3/8 故sin³(π+α)+cos³(2π-α) =[-sina]³+[cos(-a)]³ =-sin³a+cos³a =cos³a-sin³a =(cosa-sina)(cos²a+sinacosa+s...

解: A、 sin15°cos15°=1/2sin30°=1/4, B、 cos²(π/12)-sin²(π/12)=cos(π/6)=√3/2, C、 sin12°cos42°-cos12°sin42°=sin(12°-42°)=sin(-30°)=-1/2, D、 tan22.5°/(1-tan²22.5°) =sin22.5°/cos22.5°/[(cos²22...

网站首页 | 网站地图
All rights reserved Powered by www.mpjx.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com