sin(90度-角a)=-搜答案-搜搜考试网

(tanα∙sinα)/(tanα-sinα)=(tanα∙sinα)/[sinα(1/cosα-1)]=tanα/[(1/cosα)-1)]=(tanα∙cosα

正确,设直角三角形ABC所对应的三边分别为a,b,c,则由勾股定理知a^2+b^2=c^2sin平方A+sin平方B=(a/c)^2+(b/c)^2=(a^2+b^2)/c^2=c^2/c^2=1c^

sin(45度-a)=5/13sin(25pai/4+a)=sin(pai/4+a)=cos[90°-(45°+a)]=cos(45°-a)=±√[1-sin²(45°-a)]=±12/13

sin20*sinA=sin80*sin(A-20)2sin10cos10*sinA=cos10*sin(A-20)2sin(30-20)sinA=sin(A-20)2sinAsin30cos20-2

sin^2A=sin^2B+sin^2C+sinBsinC则有a^2=b^2+c^2+bc,bc=-(b^2+c^2-a^2)cosA=(b^2+c^2-a^2)/2bc=-1/2=cos120,A=

因为sin^2A+cos^2A=1cos2A=cos^2A-sin^2A所以cos^2A=(cos2A+1)/2sin^2A=(1-cos2A)/2因为A+B=90°所以2A+2B=180°所以cos

∵a是第三象限角∴180°

sinC=sin(A+B),给你一点提示,把A+B看做整体,再证证看.原式1,为你能想到什么?

左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin²a(1-sin

sin(15度-a)=1/5a是第一象限角cos(15度-a)=2√6/5sin(165度+a)-cos(165度+a)=sin(15度-a)+cos(15度-a)=(1+2√6)/5再问：cos(1

好像做错了,哈哈

-(M-1)=-M+1

4tan(a/2)=1-tan(a/2)得2tan(a/2)/(1-tan(a/2))=1/2tan(a)=1/24sinb=sin(2ab)sinb=2sin(ab)cos(a)2sinb=sin(

求证sin(90+a)=cos a

sin(90+a)=sin90*cosa+cos90*sina=cosaen

直角三角形中,AC=√(12*12+5*5)=13sinA=BC/AC=5/13cosA=AB/AC=12/13tanA=BC/AB=5/12

x垂直于y=>x.y=0=>(a+(t^2-3)b).(-ka+tb)=0-k|a|^2+t(t^2-3)|b|^2+(t-k(t^2-3))a.b=0-4k+t(t^2-3)+(t-k(t^2-3)

设直角三角形ABC三个角所对边分别是：a,b,c则：a²+b²=c²所以：sin²A+sin²B=(a/c)²+(b/c)²=[a

Sin[a+b]Sin[a-b]积化和差公式得=1/2(-Cos[2a]+Cos[2b])余弦二倍角公式得=Sin[a]^2-Sin[b]^2Sin[80°]Sin[40°]=Sin[60°+20°]

应该是sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]A=(A+B)/2+(A-B)/2.B=(A+B)/2-(A-B)/2所以sin(A+B)/2cos(A-B)/2+cos(

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