已知sin(四分之π x)=-五分之四

∵sin(π/4-x)=12/13∴cos(π/4+x)=cos(π/2-(π/4-x))=sin(π/4-x)=12/13∵sin(π/4-x)=(cosx-sinx)/√2∴cosx-sinx=√

有sin(x-45°)=√2／4=sinxcos45°-cosxsin45°,得sinx-cosx=0.5,两边平方得1-2sinxcosx=0.25.sinxcosx=3／8.tanx+1／tanx

四分之三X+四分之一=八分之五两边乘86x+2=56x=5-26x=3x=3/6x=1/2x-3/4x=181/4x=18x=18x4x=72

f(x)=sin2x+2sin（π/4-x）cos（π/4-x）=sin2x+cos2x=√2sin（2x+π/4）所以T=2π/2=πx属于[-π/12,π/2]得到2x+π/4属于[π/12,5π

∵0＜x＜π/4∴0再问：求的是cos2x除以cos（四分之派加x)的值，你算的是cos2x除以cos（四分之派减x)，算错了，正确答案是13分之24，帮忙再算一下，谢谢！再答：啊我算的是cos2x/

解；f(x)=sinx+sinxcosπ/3+cosxsinπ/3=sinx+1/2sinx+√3/2cosx=3/2sinx+√3/2cosx=√3sin(x+π/6)当x+π/6=-π/2+2kπ

f（x）=-1+cos方x-cosx+9/4=cos方x-cosx+5/4=(cosx-1/2)^2+1cosx=1/2,f(x)最小=1cosx=-1,f（x）最大=9/4+1=13/4有问题留言

(1)cosa=-√1-sina^2=-12/13,(2)sin(兀/4-a)=sin(兀/4)cosa-cos(兀/4)sina=-17√2/26

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=sinx*cos(π/6)+cosx*sin(π/6)+sinx*cos(π/6)-cosx*sin(π/6)+cosx+a=（√3

π/2

x+y=10x-y=8/7原式=-7³/8*(x+y)³*(x-y)*4(x+y)²(x-u)²=-343/2*(x+y)^5*(x-y)³=-343

你好!数学之美团为你解答α,β∈(3π/4,π)α+β∈(3π/2,2π),β-π/4∈(π/2,3π/4)sin(α+β)=-3/5,sin(β-π/4)=12/13∴cos(α+β)=4/5,co

等一下再答：

∵0＜a＜四分之π＜β＜四分之三π∴3π/4

4x/7=5(x-6)/835(x-6)=32x35x-35*6=32x3x=35*6x=70

17π/12＜x＜7π/4,得5π/3＜x+π/4＜2πcos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)

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

根据A是锐角,那么A∈(0,π/2)所以A-π/4∈(-π/4,π/4)另外根据sin(A-π/4)=√5/5>0所以A-π/4∈(0,π/4)所以cos(A-π/4)=2√5/5所以sinA=sin

解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)

由题意,tana=2或1/2,又a∈[π/4,π/2],tana>1,故tana=2由万能公式,cos2a=1-tan^2a/1+tan^2a=1-4/1+4=3/5sin2a=2tana/1+tan