已知sin(4分之π-x)=13分之5,0

sin(PI/6+2x)=cos(PI/2-PI/6-2x)=cos(PI/3-2x)=cos(2*(PI/6-x))=1-2*sin(PI/6-x)^2=1-2*(1/4)^2=7/8tan70*c

先用tanx=sinx/cosx、倍角公式、诱导公式化简原函数：f（x）=sin²x+sinxcosx-sin[2（x+π/4）]=（1-cos2x）/2+1/2sin2x-sin（2x+π

(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

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

因为sin(4分之π-x)等于13分之5,即sin（π/4-x）=5/13cos(x-π/4+π/2)=-sin(x-π/4)=sin(π/4-x)=5/13cos[2(x+π/4)]=-sin2x=

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

公式忘了,就凑乎着做吧1.（1+cotx）sin^2(x)-2sin(x+π／4)sin(x-π／4)=(1+cosx/sinx)sin^2(x)-(cosπ／2-cos2x)=sin^2(x)+co

解题思路：同学你好，你给的题目，条件不明确。我解答时，添加的条件是【若该方程在区间[-π/6,13π/12]上面有解】，若题目还有其他情况，请添加讨论.解题过程：

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

∵cos（π/4+x）=4/5,∴cosx-sinx=(4/5)√2两边平方得,cos²x+sin²x-2sinxcosx=32/25则,2sinxcosx=-7/25∴（cosx

sin（x-4分之π）＝－sin（π/4-x)=-sin[π/2-(π/4+x)]=-cos[(π/4)+x]=1/3所以：cos[(π/4)+x]＝－1/3

f(5π/4)=2sin(5π/12-π/6)=2sin(3π/12)=2sin(π/4)=√2

最大值:f(π/8)=(√2)sin(π/4+π/4)=(√2)sin(π/2)=√2最小值:f(-π/4)=(√2)sin(-π/2+π/4)=(√2)sin(-π/4)=√2*(-√2/2)=-1

将函数f(X)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π/4)化简得：=(1+cosx/sinx)*2sinxcosx+m(sinxcosπ/4+cosxsinπ/4)(

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

f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

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

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)