f(x)=4cos^4x-2cos2x-1/sin(x+45)sin(x-45) ,求f(-11π/12)的值

f(x)=（4cos^4x-2cos2x-1）/[sin(π/4+x)·sin(π/4-x)]
=[(1+cos2x)*2-2cos2x-1]/[(-1/2)(cosπ/2-cos2x)]
=2(cos2x)^2/cos2x=2cos2x
1)f(（-11/12)π）=2cos(-11π/6)=2cos(11π/6)=2cos(2π-π/6)=2cos(π/6)=根号3
2)g(x)=1/2f(x)+sin2x=cos2x+sin2x=根号(2)sin(2x+π/4),
当2x+π/4=π/2,x=π/8时,g(x)最大=根号2,当x=0时,g(x)最小=g(0)=1.

原式=4(sin^2x+cos^2x)^2-8sin^2xcos^2x-3
=1-8sin^2xcos^2x
=1-2(sin2x)^2
=cos4x

分子化简下，4cos^4x-2cos2x-1=（2cosx＾2）＾2-2cos2x-1=(1+cos2x)＾2-2cos2x-1=cos2x＾2
分母其实不化简也行，直接代入都可以算了，tan（π/4+x)=tan(3π/12-17π/12)=tan(-7π/6)=tan(5π/6)=-tan(π/6)=-√3/3，sin^2(π/4-x)]=sin^2(π/4+17π/12)=sin^2...

y=7-4sinxcosx+4cos^2x-4cos^4x
cos^2x=(cos2x+1)/2
cos^4x=(cos2x+1)^2/4
sinxcosx=(sin2x)/2
所以y=7-2sin2x+2cos2x+2-(cos2x+1)^2
=7-2sin2x+2cos2x+2-(cos2x)^2-2cos2x-1
=8-2sin2x-(cos2x)^2
=8-2sin2x-[1-(sin2x)^2]
=(sin2x)^2-2sin2x+7
=(sin2x-1)^2+6
-1

f(x)=(4cos^4x-2cos2x-1)/[sin(π/4+x)*sin^2(π/4-x)]
=(4cos^4x-4cos²x+1)/[sin(π/4)*cosx+sinx*cos(π/4)*(1-cos(π/2-2x))/2]
=(2cos²x-1)²/[√2/2*(cosx+sinx)*(1-sin2x)/2]
=(cos²x-sin²x)²/[√2/4*(cosx+sinx)*(cosx-sinx)²
=(cosx+sinx)²*(cosx-sinx)²/[√2/4*(cosx+sinx)*(cosx-sinx)²
=2√2*(cosx+sinx)=4*sin(π/4+x)

因为4cos^4x=(2cos^2x)^2=(1+cos2x)^2=1+2cos2x+(cos2x)^2
所以易得f(x)=cos2x
(1)f(-11/12π)=cos(-11/6π)=cos(2π-π/6)=-cos(π/6)=(负根号3）/2
(2)g(x)=f(x)+sin2x=（根号2）sin(2x+π/4)
因为x属于[0,π/4] 所以(2x+π/4)属于【π/4,3π/4】所以sin(2x+π/4)属于[根号2/2,1]
g(x)max= （根号2） g(x)min=1

1.f(x)=[(2cos²x)²-2cos2x-1]/cos2x
=[(1+cos2x)²-2cos2x-1]/cos2x
=(cos2x)²/cos2x
=cos2x
(1)f(-11/12π)
=cos(-11/6π)
=cos(11/6π)
=cos(2π-π/6)
=cosπ/6=√3/2
(2)g(x)=√3f(x)+sin2x
=√3cos2x+sin2x
=2[(√3/2)cos2x+(1/2)sin2x]
=2cos(2x-π/6)
x∈[0,π/4]时
2x-π/6∈[-π/6,π/3]
所以,
x=π/12时,g(x)最大,为2
x=π/4时,g(x)最小,为1
2
在等比数列{an}中,已知a2=2,a5=6
所以,q³ = 6/2 = 3
所以,q=三次根号下3
an = 2*(q)^(n-2)
若等差数列{bn}中,b1=a5=6,b8=a2=2,
所以,7d = 2-6=-4
所以,
d = -4/7
b1 = 6
所以,
Sn = 6n + n(n-1)d/2
= 6n - 2(n²-n)/7
= (-2/7)n² + 44/7n
所以对称轴为n = 11
n = 11时,Sn最大 = 242/7

y=7-4sinxcosx+4cos²x-4cos^4x
=7-2*2sinxcosx+4cos²x(1-cos²x) (提公因式4cos²x)
=7-2*2sinxcosx+4cos²x*sin²x ( 同角三角函数的平方公式：sin²x+cos²x=1)
=7-2*2sinxcosx+(2sinxcosx)²
=7-2sin2x+sin²2x (二倍角公式：sin2x=2sinxcosx)
=(sin²2x-2sin2x+1)+6
=(sin2x-1)²+6 ( 完全平方公式：a²-2ab+b²=(a-b)² )
也即y=(1-sin2x)²+6

f(x)=4cos^4x+4cos^2x-3/cos2x=4cos^4x+4cos^2x-3/(2cos^2x-1)
因为2cos^2x-1不为0,故定义域为{xx不等于土pi/4+2k*pi,k为整数)
令cos^2x=t,t属于[0,1)
那么原式为f(t)=4t^2+4t-3/(2t-1)
求导,f'(t)=8t+4+3/(2t-1)^2
因为t属于[0,1),故f'(t)恒大于0,所以f(t)单调递增,其值域为[3,5)

f(x)=1-4sinxcosx+4cos^2x-4cos^4x是f(x)=1-4sinxcosx+4(cosx)^2-4(cosx)^4这个吧? f(x)=1-4sinxcosx+4(cosx)^2[1-(cosx)^2]=1-4sinxcosx+4(sinxcosx)^2=（2sinxcox-1)^2 =(sin2x-1)^2 -1=

f(-11π/12)=4cos^4(11π/12)-2cos(11π/6)+1/sin(7π/6)sin(2π/3)
sin2π/3sin(7π/6)=-sinπ/6sinπ/3=-√3/4
2cos11π/6=2cos(2π-π/6)=2cosπ/6=√3
4cos^4x=(2cos^2x)^2=(1+cos2x)^2
cos2x=cos22π/12=√3/2
(1+√3/2)^2=(1/4)(4+2√3)=1+√3/2
f(-11π/12)=1+√3/2-√3-4/√3
=1-11√3/6