根号2sin(2x-4分之π)的增区间

f(x)=√2*(2x-π/4)/sin(x+π/2)∵cosa=3/5∴a=arccos3/5f(a)=√2*(2a-π/4)/sin(a+π/2)=√2*(2arccos3/5-π/4)/cosx

cosx=1/7cos(x-y)=13/14siny=sin(x-(x-y))=sinxcos(x-y)-cosxsin(x-y)=4√3/7*13/14-1/7*3√3/14=√3/2所以y=π/3

利用三角函数积化和差公式sinAcosB=0.5[sin(A+B)+sin(A-B)]和倍角公式cos2A=cos²A-sin²A原式=sin²x+√3[sin2x+si

sqrt(2)*cos（2x-π/4）=cos(2x)+sin(2x)sin(x+pi/2)=-cos(x)1+cos(2x)=2*cos^2(x)sin(2x)=2*sin(x)*cos(x)所以最

2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[

是四份之π还是π分之四?再问：四分之π再答：你的题给的有点特殊，所以只能按照我理解的给答案，如果有错可以追问f(x)=ab={√2sin(π/4+x)+1}x{√2sin(π/4+x)-1}-√3co

(1)首先利用降幂公式：f(x)=1-cos(π/2+2x)-根号3cos2x再利用诱导公式：f(x)=1+sin2x-√3cos2x最后是辅助角公式:f(x)=2*(1/2*sin2x-√3/2*c

f(x)的定义域为全体实数另g(x)=sin(2x+3π/2)f(x)与g(x)奇偶性一致（f(x)/g(x)=根号2>0）g(x)=sin(2x+3π/2)=-cos(2x)g(-x)=-cos(-

√2sin(x-π/4)=√2(sinxcosπ/4-cosxsinπ/4)=√2[(1/√2)sinx-(1/√2)cosx]=sinx-cosx

已知函数f(x)=2倍根号3sin(x-4分之π)cos(x-4分之π)-sin(2x-π)（1）求函数f（x）的单调递减区间2）试试画出函数f（x）在0到π上的图像(1)解析：∵函数f(x)=2倍根

f(x)=2sinπ/4cosπ/4-2√3sin²π/4+√3=sinπ/2+√3(1-2sin²π/4)=1+√3cosπ/2=1

化简后得（cosa）²-(sina)²／sinacosπ/4-cosasinπ/4=-√2/2（cosa+sina）（cosa-sina）/√2/2（sina-cosa）=-√2/

f'(x)=2√2cos(2x+π/4)当2nπ-π/2

经济将积极

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3)

f(x)=[1-√2sin(2x-π/4)]/cosx=[1-√2(sin2xcosπ/4-cos2xsinπ/4)]/cosx=[1-√2(√2/2*sin2x-√2/2*cos2x)]/cosx=

解:f(x)=1+√2sin(2x-π/4).1.函数f(x)的最小正周期为T=2π/2=π.当思念(2x-π/4)=1时,函数f(x)具有最大值,且f(x)min=1+√2.2.函数的增区间:∵si