3sin(x 10°) 5sin(x 70°)的最大值
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sin(5/6兀)+sin的平方(兀/3-x)=-sin(5/6π-π)+cos²(π/2-π/3+x)=-sin(-π/6)+cos²(π/6+x)=sin(π/6)+1-sin
y=3sin(x+20°)+5sin(x+80°)=3sin(x+20°)+5sin(x+20°+60°)=3sin(x+20°)+5[sin(x+20°)cos60°+cos(x+20°)sin60
3(sinA)^2+2(sinB)^2=5sinA(sinA)^2+(sinB)^2=5sinA/2-(sinA)^2/25sinA/2-(sinA)^2/2=-(1/2)(sinA-5/2)^2+2
直接画不行吗t=0:5:600;y=sin(314*t)+sin(3*314*(t-0.065))+sin(5*314*(t-0.09))+sin(11*314*(t-0.14));plot(x,y)
sinα+cosβ=3/4,cosα+sinβ=-5/4(sinα+cosβ)^2=9/16(cosα+sinβ)^2=25/16上两式相加2+2sin(α+β)=34/16sin(α+β)=1/16
答案是:2/5.令A=(α+β)/2,B=(α-β)/2,则有:2*sinA*cosA=2/3,2*sinB*cosB=3/5,2*sinA*cosB=1/2要求的是:cosA*sinB=(2/3)*
我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积分限是0→π∫[0→π]√(sin^3x-sin^5x)dx=∫[0→π]√[sin³x(1-sin²
sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π
先积化和差sin3xsin5x=0.5(cos2x-cos8x)∫sin3xsin5xdx=∫0.5(cos2x-cos8x)dx=0.25sin2x-0.0625sin8x+c
x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)
3sina+cosa=03sina=-cosatana=sina/cosa=-1/31、(3cosa+5sina)/(sina-cosa)【分子分母同除以cosa】=[3+5tana]/[tana-1
[3/2,13/4]
cos[(α+β)/2]*sin[(α-β)/2]=(1/2)·(sinα-sinβ)(用积化和差公式,或把乘式的每一部分按两角和差的正,余弦展开求出);sin(α+β)=sinαcosβ+sinβc
sin(2α+β)=5sinβsin[α+(α+β)]=5sin[(α+β)-α]sinαcos(α+β)+cosαsin(α+β)=5sin(α+β)cosα-5cos(α+β)sinα6sinαc
tga=5√3/11,0
本题题目应是要证:2tan(α+ β)=3tanα,答案见图片:
已知(2sinθ+cosθ)/(sinθ-3cosθ)=-5,求3cos2θ+4sin2θ的值∵(2sinθ+cosθ)/(sinθ-3cosθ)=-5∴(2tanθ+1)/(tanθ-3)=-5,解
sin²1°+sin²2°+sin²3°...+sin²45°+sin²46°...+sin²89°=sin^2(90-89)+sin^2(
答:sin^2a+sin^2(a+60)+sin^2(a+120)=3/2.证明:左边=sin^2a+sin^2(a+60)+sin^2(a+120)=sin^2a+(sinacos60+cosasi
sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2