三道关于高一三角函数的题目1)求值cos(π/5)cos(2π/5)2)求函数y=1-(cosx)^2+(cosx)^4的最小正周期3)求函数y=2sinx(sinx+cosx)的减区间1)0.25 2)π/4 3)[kπ+3/8π,kπ+7/8π]
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三道关于高一三角函数的题目1)求值cos(π/5)cos(2π/5)2)求函数y=1-(cosx)^2+(cosx)^4的最小正周期3)求函数y=2sinx(sinx+cosx)的减区间1)0.25 2)π/4 3)[kπ+3/8π,kπ+7/8π]
三道关于高一三角函数的题目
1)求值cos(π/5)cos(2π/5)
2)求函数y=1-(cosx)^2+(cosx)^4的最小正周期
3)求函数y=2sinx(sinx+cosx)的减区间
1)0.25 2)π/4 3)[kπ+3/8π,kπ+7/8π]
三道关于高一三角函数的题目1)求值cos(π/5)cos(2π/5)2)求函数y=1-(cosx)^2+(cosx)^4的最小正周期3)求函数y=2sinx(sinx+cosx)的减区间1)0.25 2)π/4 3)[kπ+3/8π,kπ+7/8π]
1)1/4
∵cos(π/5)cos(2π/5)+sin(π/5)sin(2π/5)=cos(π/5)
cos(π/5)cos(2π/5)-sin(π/5)sin(2π/5)=cos(3π/5)
∴cos(π/5)cos(2π/5)=[cos(π/5)+cos(3π/5) ]/2
又∵cos(π/5)+cos(3π/5)=cos(π/5)-cos(2π/5)
cos(π/5)-cos(2π/5)=-cos(4π/5)-cos(2π/5)
cos(π/5)-2cos(π/5)^2+1=-cos(2π/5)-2cos(2π/5)^2+1
cos(π/5)-2cos(π/5)^2=-cos(2π/5)-2cos(2π/5)^2
cos(π/5)+cos(2π/5)=[cos(π/5)^2-cos(2π/5)^2]*2
[cos(π/5)-cos(2π/5)]*2=1
∴cos(π/5)cos(2π/5)=[cos(π/5)+cos(3π/5) ]/2=1/4
2）y=1-(cosx)^2+(cosx)^4
=1+(cosx)^2*[(cosx)^2-1]
=1+(sin2x)^2/4
3）y=2*(sinx)^2+sin2x
=1-cos2x+sin2x
=1- (根号2)*sin(x-π/4)
汗~~~，这也问