www.fltk.net > 求证Cos7x+7Cos5x+21Cos3x+35Cosx=64Cos^7x

求证Cos7x+7Cos5x+21Cos3x+35Cosx=64Cos^7x

cos7x=cos5x cos7x - cos5x = 0 -2 sin6x sinx = 0 sin6x = 0,或者 sinx = 0 6x = kπ,或者 x = kπ x=(k/6)π,或者 x = kπ x =(k/6)π k为整数

证明 分子 sin(4x+3x)+sin(4x-3x) =sin4xcos3x+cos4xsin4x+sin4xcos3x-cos4xsin4x =2sin4xcos3x 分母 sin(4x+x)+sin(4x-x) =sin4xcosx+cos4xsinx+sin4xcosx-cos4xsinx =2sin4xcosx 左=(2sin4xcos3x)/(2sin4xcosx)=cos3x/cosx,其中 ...

∫ (7 cos 5x -5sin7x)dx=7/5sin5x+5/7cos7x+C

三角函数积化和差公式:cosAxcosBx=(1/2)[cos(A+B)x+cos(A-B)×],当A=6、B=1时为:cos6xcosx=(1/2)(cos7x+cos5x),将此式代入原式左端积分号后再将系数1/2提在前面即得右端形式。

用分部积分法两次:

解:∫sin5Xsin7Xdx =∫-(1/2)[cos(5x+7x)-cos(5x-7x)]dx =-(1/2)∫(cos12x-cos2x)dx =(-1/2)[(1/12)sin12x-(1/2)sin2x]+C =-(1/24)sin12x+(1/4)sin2x+C

积化和差:cosacosb=1/2*[cos(a+b)+cos(a-b)] cos5xcos7x=1/2*[cos(5+7)x+cos(5-7)x]=1/2*[cos12x+cos2x] ∴∫cos5xcos7xdx =1/2∫(cos12x+cos2x)dx =1/2∫cos12xdx+1/2∫cos2xdx =1/2*1/12*∫cos12xd(12x)+1/2*1/2*∫cos2xd(2x) =1/24*sin12x+1/4*sin2x+C

sin5xsin7x=-2分之1[cos(5x+7x)-cos(7x-5x)]=-2分之1(cos12x-cos2x) sin5xsin7x的不定积分=-24分之1sin12x+4分之1sin2x+c

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