怎么证明sin²a+cos²a=1

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

tana-cota＝（切化弦）（sina^2-cosa^2）/sinacosa=(sina+cosa)(sina-cosa)/sinacosaseca+csca=(sina+cosa)/sinacos

(sina+tana)(cosa+cota)=(sin(a)+sin(a)/cos(a))/(cos(a)+cos(a)/sin(a)),展开=sin(a)*cos(a)+cos(a)+sin(a)+

证明：输入过于麻烦,用换元法吧设A=sin²A,B=sin²B∵sin^4a/sin^2b+cos^4a/cos^2b=1即A²/B+(1-A)²/(1-B)=

不知道你的题目是否有错：(cosa-sina+1)/(cosa+sina+1)=(1-sina)/cosa证明－：设点P（X,Y）是角a终边上的任意一点,则sina=Y/[(x^2+y^2)^(1/2

Sin^2A+Sin^2B-Sin^2ASin^2B+Cos^2ACos^2B=Sin^2A(1-Sin^2B)+Sin^2B+Cos^2ACos^2B=Sin^2ACos^2B+Cos^2ACos^

根据余弦2倍角公式cos（a+b）=cosa*cosb-sina*sinbcos2a=cos（a+a）=cosa*cosa-sina*sina=cos²a-sin²a再根据三角函数

分子=sin(n+1)A+2sin(n)A+sin(n-1)A=[sin(n+1)A+sinnA]+[sinnA+sin(n-1)A]=2sin(2n+1)A/2*cosA/2+2sin(2n-1)A

右边=2sin(a/2)cos(a/2)/(1+2cos(a/2)平方-1)=sin(a/2)/cos(a/2)=左边

tana/2=(1-cosa)/sina证：等式右边=[2cos^2a/2]/2sina/2cosa/2=tana/2sina/(1+cosa)=(2sina/2cosa/2)/(2-2sin^2a/

cos(A+B)的平方-sin(A-B)的平方=[cos(A+B)-sin(A-B)][cos(A+B)+sin(A-B)]=[cosAcosB-sinAsinB-sinAcosB+cosAsinB]

sina=sin[(a＋b)/2＋(a－b)/2]=sin(a＋b)/2cos(a－b)/2＋cos(a＋b)/2sin(a－b)/2sinb=sin[(a＋b)/2－(a－b)/2]=sin(a＋b

（cosa-sina+1）/（cosa+sina+1）=（1-sina）/cosa注意1+cosa=2(cosa/2)^2,sina=2sina/2cosa/2左边=[2(cosa/2)^2-2sin

http://hi.baidu.com/wolf224/blog/item/296fec120385450a5aaf53c7.html

=sin4Asin2A+cos4Acos2A-cos2A(cos4Acos2A-sin4Asin2A)=cos2A+cos2Acos6A=cos2A(1+cos6A)

∵cosx=(e^ix+e^-ix)/2e^ix=cosx+isinx∴cos(a+b)=[e^i(a+b)+e^-i(a+b)]/2=[(cosa+isina)(cosb+isinb)+(cosa-

三角函数定义sina=y/rcosa=x/rr^2=x^2+y^2sin^2a+cis^2a=(y/r)^2+(x/r)^2=(x^2+y^2)/r^2=1

证明：sina+cosa=√2sin(a+∏/4)由于|sin(a+∏/4)|≤1;所以|sina+cosa|=√2|sin(a+∏/4)|≤√2即sina+cosa∈[-√2,√2]再问：为什么si

sinA=a/ccosA=b/csin平方A+cos平方A=(a/c)^2+(b/c)^2=【a^2+b^2】/c^2=1再问：【a^2+b^2】/c^2=1为什么就等于1呢？再答：勾股定理呀