sin^4(10) sin^4(50) sin^4(70)

(3sinα-2cosα)/(5sinα+4cosα)(分子分母同时除以cosα)=(3tanα-2)/(5tanα+4)=(-2*3-2)/(-2*5+4)=-8/(-6)=4/3sin^2α+2s

cosA+cosB+cosC=2cos[(A+B)/2]cos[(A-B)/2]+cosC=2cos[(π-C)/2]cos[(A-B)/2]+cosC=2sin(c/2)cos[(A-B)/2]+1

那个前半括号里面相加等于一

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

∵4sinα−2cosα5cosα+3sinα=4tanα−25+3tanα=10，∴tanα=-2．故答案为：-2

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

分子=1-sin^6a-cos^6a=1-（sin^6a+cos^6a）=1-（sin²a+cos²a)(sin^4a-sin^acos6a+cos^4a)=1-(sin^4a-s

前面sinα-cosα×tanα=0后面-sin^4α-sin^2α×cos^2α-cos^2α=-sin⁴α-sin²α(1-sin²α)-cos²α=-s

由a范围则cos(a-π/4)>0sin²+cos²=1所以cos(a-π/4)=7√2/10cosa=cos(a-π/4+π/4)=cos(a-π/4)cosπ/4-sin(a-

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)

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

（sinx）^4+（cosx）^4=1即(sinx)^4+(cosx)^4+2(sinx)^2(cosx)^2-2(sinx)^2(cosx)^2=1即[(sinx)^2+(cosx)^2]^2-2(

sin^2A+sin^2B+sin^2C=(1-cosA)/2+(1-cosB)/2+(1-cos^2C)=2-cos(A+B)cos(A-B)-cos^2C=2+cosCsoc(A-B)-cos^2

根据正弦定理：a/sinA=b/sinB=c/sinC=2R,R为该三角形外接圆半径,则：a/2R=sinAb/2R=sinBc/2R=sinC因此：sinA:sinB:sinC=a:b:c=3:2:

令A=arccos(-1/3)B=arcsin(-1/4)那么有cosA=-1/3sinB=-1/4由arccos,arcsin的取值范围可知,A为钝角（90°-180°之间）B为一个负的锐角（-90

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x

f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

1

已知(2sinθ+cosθ)/(sinθ－3cosθ)＝－5,求3cos2θ+4sin2θ的值∵(2sinθ+cosθ)/(sinθ－3cosθ)＝－5∴(2tanθ+1)/(tanθ－3)＝－5,解

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x