已知 cos(x Π 4)=4 5,x∈(-1 4,0),则sinx=

f(x)=cos(-2x+a)(-π

f(x)=cos^2x+sinxcosx=(1+cos2x)/2+1/2*sin2x=1/2+1/2(cos2x+sin2x)=√2/2*(√2/2*cos2x+√2/2sin2x)+1/2=√2/2

f(x)=[1+cos2x)/2]/[1+(1-cos2x)/2]=(1+cos2x)/(3-cos2x)=-1+4/(3-cos2x)f'(x)=-4/(2-cos2x)^2*(2-cos2x)'=

-cosxtanx再问：根据上题求f（-31π/3）的值再答：上面的答案可化简为-sinxf（-31π/3）=f（-π/3）=-1/2

根据公式：COS^2a=(1+COS2a)/2a=(X+π/12)说句不好听的,你还是基本知识没掌握好,不会活用知识.希望你多背多看,看清题,把公式活用.

f(x)=√2sin（2 x-π/4）+2f(x)max=√2+2；此时X=3π/8令（2 x-π/4）∈【-π/2,π/2】,解得x∈【-π/8,3π/8】,因为x∈(0&nbs

证明：2f(x)=（4cos^4x—4cos²x+1）/（2tan（π/4—x）cos²（π/4-x））=[2cos^2(2x)-1]^2/(2sin（π/4—x）cos（π/4-

解（1）因为f(x)=2cos(2x-π4)．所以函数f（x）的最小正周期为T=2π2=π，由单调区间-π+2kπ≤2x-π4≤ 2kπ，得到-3π8+kπ≤x≤π8+ kπ故函数

f(x)=2sin(x+π/4)cos(x-π/4)=2sin(x+π/4)cos(π/4-x)=2sin(x+π/4)cos[(π/2)-(x+π/4)]=2sin²(x+π/4)=1-c

因为π/4+x和π/4-x是互余的（和是π/2）,所以cos(π/4+x）=sin(π/4-x)=3/5因为0＜x＜π/4所以sin(π/4+x)=4/5(因为π/4+x为第一象限角)

f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+

f(x)=cos(π/3+x)cos(π/3-x),=(cosxcosπ/3-sinxsinπ/3)(cosxcosπ/3+sinxsinπ/3)=(1/2cosx-√3/2sinx)(1/2cosx

f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1

1）利用2倍角公式f(x)=sinx+sin[2(π/4+x/2)]=sinx+sin(π/2+x)=sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(cosπ/4sinx+sin

cos2x=3/5,sin2x=±4/5sin^4x+cos^4x=（-sinx^2+cosx^2)^2+2sinx^2cosx^2=cos2x^2+1/2sin2x=9/25±2/5=19/25,或

f(x)=sin(2x)+√2cos(x-π/4)=sin(2x)+√2[cosxcos(π/4)+sinxsin(π/4)]=sin(2x)+cosx+sinx=sin(2x)+√2sin(x+π/

把钱抢回来!ω=2φ=1/2

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

根据题意可知：sinx>0cosx

f(x)=sin²(π/4+x)+cos²x+1/2=1/2-cos(π/2+2x)/2+cos²x+1/2=1/2+sin2x/2+1/2+cos2x/2+1/2=si