求值cos(pai/4-x)cos(pai/4+x)=1/4 求sin^4x+cos^4x
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/23 23:34:42
求值cos(pai/4-x)cos(pai/4+x)=1/4 求sin^4x+cos^4x
要详细过成,我没有分了,
要详细过成,我没有分了,
sin^4x+cos^4x
=(sin^2x + cos^2x)^2 - 2*sin^2x*cos^2x
=1 - 2*sin^2x*cos^2x
为避免混淆,还是把乘方写在外面吧
= 1 - 2*(sinx)^2*(cosx)^2
= 1 - 2*(sinx*cosx)^2
= 1 - (sin2x)^2/2
而另一方面
cos(pai/4-x)cos(pai/4+x)= 1/4 =
[cos(pai/4)*cosx + sin(pai/4)sinx)][cos(pai/4)cosx - sin(pai/4)sinx]
=[cos(pai/4)cosx]^2 - [sin(pai/4)sinx]^2
=[(cosx)^2 - (sinx)^2]/2
=cos2x/2
所以 cos2x = 1/2
sin2x = ±√[1-(cos2x)^2 = ±√3/2
(sin2x)^2 = 3/4
因此
(sinx)^4 + (cosx)^4
= 1 - (sin2x)^2/2
= 1 - (3/4)/2
= 1 -3/8
= 5/8
=(sin^2x + cos^2x)^2 - 2*sin^2x*cos^2x
=1 - 2*sin^2x*cos^2x
为避免混淆,还是把乘方写在外面吧
= 1 - 2*(sinx)^2*(cosx)^2
= 1 - 2*(sinx*cosx)^2
= 1 - (sin2x)^2/2
而另一方面
cos(pai/4-x)cos(pai/4+x)= 1/4 =
[cos(pai/4)*cosx + sin(pai/4)sinx)][cos(pai/4)cosx - sin(pai/4)sinx]
=[cos(pai/4)cosx]^2 - [sin(pai/4)sinx]^2
=[(cosx)^2 - (sinx)^2]/2
=cos2x/2
所以 cos2x = 1/2
sin2x = ±√[1-(cos2x)^2 = ±√3/2
(sin2x)^2 = 3/4
因此
(sinx)^4 + (cosx)^4
= 1 - (sin2x)^2/2
= 1 - (3/4)/2
= 1 -3/8
= 5/8
f x =sin(pai*x/4-pai/6)-2(cos pai*x/8)^2+1
f(x)=(1+cos2x)/[4sin(pai/2+x)]-asin(x/2)cos(pai-x/
f(x)=cos(2x+pai/4)+sin(2x+pai/4)求单调区间
已知函数f(x)=cos(2x-pai/3)+2sin(x-pai/4).sin(x+pai/4)求函数在区间[-pai
已知sin(x-pai/4)=根号2/10,X属于(pai,5pai/4) 求sinx得值.求cos(2x-pai/6)
cos(x-pai/4)=根号2/10,x属于(pai/2,3pai/4)求sinx和sin(2x+pai/3)
已知:cos(x-pai/4)=根号2/10,x属于pai/2,3pai/4,求sin(2x+pai/3)的值
化简:1,sin(pai/4-x)sin(pai/4+x) 2,cosA+cos(120度-A)+cos(120度+A)
已知cos(x-pai/4)=根号2/10,x€(pai/2,3pai/4).(1)求sinx的值(2)求sin(2x+
f(x)=[1+根号2cos(2x-pai/4)]/[sin(x+pai/2)]求定义域 (2)若角a在第一象限且cos
已知函数f(x)=1+cos2x/4sin(pai/2-x)-asinx/2cos(7pai-x/2)
已知函数f(x)=2sin(x+pai/6)-2cos,x属于[pai/2,pai].若sinx=4/5求f(x)的值,