已知cos(x-派 4)=根号2 10,x属于(派 2,3派 4
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根据已知条件,sin(π-a)-cos(π+a)=sqrt(2)/3,因为sin(π-a)=sin(a),cos(π+a)=-cos(a),所以上式等价于sin(a)+cos(a)=sqrt(2)/3
f(x)=2根号3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).=(根号3)sin(x+π/2)-sin(x+π)=(根号3)cosx+sinx这一步用到诱导公式=2*((根号3
cosa=1/7,cos(60°+a)=cos60°cosa-sin60°sina=-11/14
2)f=1/2=sin(x)=kπ+π/6or2kπ+π/62n项和,可分奇数n项和偶数n项s奇=a1n+n(n-1)d/2=π*n/6+π*n(n-1)=π(n^2-5n/6)s偶=a1n+n(n-
f(x)=(sinx)^2+2√3sin(x+π/4)cos(x-π/4)-(cosx)^2-√3=-((cosx)^2-(sinx)^2)+√3(sin(x+π/4+x-π/4)+sin(x+π/4
f(x)=2根号3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).=√3sin(x+π/2)+sinx=sinx+√3cosx=2(1/2sinx+√3/2cosx)=2sin(x
若cos(π+x)=-根号3/2,那么:-cosx=-根号3/2即cosx=根号3/2>0则可知角x是第一或第三象限角因为cos(π/6)=根号3/2,所以:cos(-π/6)=cos(π/6)=根号
x属于(派/2,3派/4),X+派/4属于(3派/4,派),所以sin(X+派/4)=7√2/10.Sinx=sin[(X+派/4)-派/4]=sin(X+派/4)cos派/4-cos(X+派/4)s
才5分==再问:提高了再答:1、π;5/2,1/22、-π/12再问:第二个问能详解一下么?谢再答:奇函数的话就意味着有一个对称中心(0,0),这时是最小的
利用两角和差公式和2倍角公式.f(x)=根2(sin2xcospi/4-cos2xsinpi/4)+cos4x=sin2x-cos2x+cos4x.如果你是高中的学生,我可以肯定的告诉你,你题目抄错了
[sinx+cos(π+x)]/[sinx+sin(π/2-x)]=(sinx-cosx)/(sinx+cosx)=(tanx-1)/(tanx+1)=1-[2/(tanx+1)]sinx=√3/3∴
cos2(x-派/4)=2cos(x-派/4)^2-1=-24/25=cos(2x-π/2)=-sin2x所以sin2x=24/25=2sinx*cosxcos2x=-7/25=1-2sinx^2si
cosx+cos(x-派/3)=2cos(2x-π/3)/2cos(x-x+π/3)/2=2cos(x-π/6)×cosπ/6=2×(-√3/3)×√3/2=-1
cos(x-π/4)=cosxcosπ/4+sinxsinπ/4=(√2/2)(cosx+sinx)=√2/10cosx+sinx=1/5(cosx+sinx)^2=(cosx)^2+2cosxsin
f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)=sin(2x+2π/3)-√3[
(1)T=2π/2=π增区间:2kπ-π≤2x-π/4≤2kπ2kπ-3π/4≤2x≤2kπ+π/4kπ-3π/8≤x≤kπ+π/8所以增区间为[kπ-3π/8,kπ+π/8]k∈Z(2)x∈[-π/
f(x)=2sin(2x-π/3),x∈[π/4,π/2]时,2x-π/3∈[π/6,2π/3],所以f(x)最大2,x=5π/12时;f(x)最小1,x=π/4时.|f(x)-a|
题设条件可求得cosθ和sinθ,平方相加利用二倍角公式进结论.证明:由已知式可得cosθ=,sinθ=.平方相加得cos2B+cos2C=sin2A∴+=sin2A∴cos2B+cos2C=2sin
因为x属于(pi/2,3pi/4),所以x-pi/4属于(pi/4,pi/2)在第一象限,所以sin(x-pi/4)=4/51.sinx=sin(x-pi/4+pi/4)=sin(x-pi/4)cos