3sin(x 10°) 5sin(x 70°)的最大值

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