cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 08:11:02
cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
xUKo@+{ÏNR MK ШJDSRP_ us_`vNygmW~g薳`= 1uBC]Z^φJ7tn<haoOQ(*+Rոsp`sMA z'IDxRZ=w P&E{?(^o - 3@O,~?4MIF:Y3qvq3X ߻`/3b);;S9OqD⣍)g`&M4|܌YU2#E{N|V~əJe|]i:5<+[eۤb΍+ mpY7om f; lz8v(iXD㗽 G3nd\+ lN,!̈́"1)B_ "4~(h dJB*ƓwA|'XH;RElH,p+fK lDu[S̿rC#Y(ir)p*IM'4",jD@ @E{

cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4)
利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.
已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2

cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
cos x + sin x = 根号2(sinπ/4cosx+cosπ/4sinx)=(根号2 )sin(π/4+x)
cos x - sin x = 根号2(sinπ/4cosx-cosπ/4sinx)=(根号2 )sin(π/4-x)

直接用和差的正弦(余弦)公式即可
cos(x-π/4)+ sin(x-π/4)
=2sqrt(2)sinx
cos(x-π/4)- sin(x-π/4)
=2sqrt(2)cosx
已知 1)cos x + sin x = 根号2, 2)cos x - sin x =根号2
两式相加(减),得
cosx=sinx=sqrt(2)/2

全部展开

直接用和差的正弦(余弦)公式即可
cos(x-π/4)+ sin(x-π/4)
=2sqrt(2)sinx
cos(x-π/4)- sin(x-π/4)
=2sqrt(2)cosx
已知 1)cos x + sin x = 根号2, 2)cos x - sin x =根号2
两式相加(减),得
cosx=sinx=sqrt(2)/2
所以x=π/4+2kπ,k是整数。
【sqrt表示根号】

收起

根据cosx+sinx=√2,√2*(√2/2*cosx+√2/2*sinx)=√2即√2*(cos(π/4)*cosx+sin(π/4)*sinx)=√2即cos(π/4)*cosx+sin(π/4)*sinx=1,所以cos(x-π/4)=1。.同理根据cos x - sin x =√2,,√2*(√2/2*cosx-√2/2*sinx)=√2即√2*(sin(π/4)*cosx-cos(π/...

全部展开

根据cosx+sinx=√2,√2*(√2/2*cosx+√2/2*sinx)=√2即√2*(cos(π/4)*cosx+sin(π/4)*sinx)=√2即cos(π/4)*cosx+sin(π/4)*sinx=1,所以cos(x-π/4)=1。.同理根据cos x - sin x =√2,,√2*(√2/2*cosx-√2/2*sinx)=√2即√2*(sin(π/4)*cosx-cos(π/4)*sinx)=√2即sin(π/4)*cosx-cos(π/4)*sinx=1,所以sin(x-π/4)=-1。所以cos(x-π/4)+ sin(x-π/4) =0,cos(x-π/4)- sin(x-π/4) =2。.

收起

cosx+sinx=√2 ,cosx-sinx=√2
cosx=√2,sinx=0
cos(x-π/4)+ sin(x-π/4)
=cosxcosπ/4+sinxsinπ/4 +sinxcosπ/4-cosxsinπ/4
=2sinxcosπ/4=0
cos(x-π/4)- sin(x-π/4)
=cosxcosπ/4+sinxsinπ/4- (sinxcosπ/4-cosxsinπ/4)
=2cosxsinπ/4=2

[(sin x-cos x)/(sin x+cos x)]^4 求定积分 积分区间0-π/4 化简1+sin x/cos x·sin2x/2cos²(π/4-x/2) 化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1 化简sin(x+7π/4)+cos(x-3π/4)步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2- 化简sin(2π+x)cos(π-x)sin(π-x)/cos(x-π)sin(-π-x)sin(π+x) sin(π/4-x)为什么等于cos(x+π/4) 化简sin(x-π/4)+cos(x+π/4) cos(x+π/4)+sin(x+π/4)化简 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). sin(sin(sin(sin(x)=cos(cos(cos(cos(x),X等于多少? tanx=m,则{sin(x+3π)+cos(π+x)}/{sin(-x)-cos(π+x)}=? 函数y=sin(x)cos(x+ π/4)+cos(x)sin(x+π/4)的最小正周期是 三角函数的计算题sin(X-2π)+sin(-x-3π)cos(X-3π)/cos(π-x)-cos(-π-x)cos(x-4π)等于多少? (1)化简:根号2sin(π/4-x)+根号6cos(π/4-x)(2)sin(67°-x)cos(x-22°)+cos(67°-x)cos(112°-x)的值为 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) sin 2x+cos 2x=根号2sin (2x+π/4)怎么得出来的, f(x)=cos⁴x+2sinxcosx-sin⁴x =(cos⁴xx-sin⁴x) +2sinxcos=(cos²x+sin²x)(cos²x-sin²x)+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)那个cos2x怎么得到的 用了什么公式还是什么定理 谢啦.