大师作文网sin2x—cos2x等于多少
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/19 23:52:31
y=sin2x-cos2x=√2sin(2x-π/4）=√2sin2（x-π/8）而y=sin2x+cos2x=√2sin(2x+π/4）=√2sin2（x+π/8）所以(x+π/8)-（x-π/8）
(1)根据图信息可知sina=4/5cosa=3/5∴(sina^2+sin2a)/(cosa^2+cos2a)=(sina^2+2sinacosa)/(cosa^2+cosa^2-sina^2)=(
f(sinx)=cos2x=1-2sin^2xf(x)=1-2x^2f(cosx)=1-2cos^2x=-cos2x选C
cos2X=(cosX)^2-(sinX)^2=2*(cosX)^2-1
sin2x=cos2xsin2x^2+cos2x^2=1∴sin2x=cos2x=根号2/2∴2x=n*pi+pi/4（n为整数）∴x=n*pi/2+pi/8
sin^2x+cos^2x=1sin^2x+9sin^2x=1sin^2x=1/10Cos2X+Sin2X=cos^2x-sin^2x+2sinxcosx=9sin^2x-sin^2x+6sin^2x
x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3
sin2x+cos2x=√2(√2/2sin2x+√2/2*cos2x)=√2(cospai/4sin2x+sinpai/4cos2x)=√2sin(2x+pai/4)所以...公式:cosasinb
sin2x+cos2x=根2·【cos45·sin2X+sin45·cos2X】=根2·sin[2X+45]
原式＝(-2cos2x/1+sin2x+cos2x)+1=(-2cos^2x+2sin^2x)/(1+2sinxcosx+cos^2x-sin^2x)+1=[2(sinx+cosx)(sinx-cos
解题思路：灵活利用三角函数的公式进行化简，最后套“周期公式”。解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prced
题目条件不充分啊cos2x+sin2x=1cos2x=cos^2x-sin^2xsin2x=2sinxcosxcos2x+sin2x=cos^2x-sin^2x+2sinxcosx
1+sin2x-cos2x=1+2sinxcosx-1+2sinx^2=2sinx(cosx+sinx)1+sin2x+cos2x=1+2sinxcosx+2cosx^2-1=2cosx(cosx+s
因为tanx=2所以tan2x=2tanx/[1-(tanx)^2]=2*2/(1-2^2)=-4/3所以(sin2x+cos2x)/(cos2x-sin2x)=[(sin2x/cos2x)+(cos
sinx/cosx=tanx所以sin2x/cos2x=tan2x
sin2x＋cos2x＝√2（√2＆＃47；2sin2x＋√2＆＃47；2＊cos2x）＝√2（cospai＆＃47；4sin2x＋sinpai＆＃47；4cos2x）＝√2sin（2x＋pai＆＃4
设f(x)=sinx^2+sin2x-2cosx^2,此题实际上就是求f(x)的值域,具体解答如下:\x0df(x)=sinx^2+sin2x-2cosx^2\x0d=(sinx^2+cosx^2)+
sinx×cos2x-sin2x×cosx=sin(x-2x)=-sinx
1/2sin4x
sinx^2+cosx^2=1,这就相当于一个公式,中间的变数x当然可以换成任何值了!