化简:
1)sin(kt-a)cos(kt+a)/sin[(k+1)t+a]*cos[(k+1)t-a]2)根号[1+cosa)/(1-cosa)]+根号[(1-cosa)/(1+cosa)]
首先补充公式 sina*cosb=(1/2)[sin(a+b)/2+sin(a-b)/2] 2sin^a=1-cos2a ,2cos^a=1+cos2a 以上是和与差的三角函数公式,二倍角公式变换而来 sin(kt-a)cos(kt+a)/sin[(k+1)t+a]*cos[(k+1)t-a] =(1/2)[sinkt-sina]/(1/2)[sin(k+1)t+sina] =[sinkt-sina]/[sin(k+1)t+sina] 2)根号[1+cosa)/(1-cosa)]+根号[(1-cosa)/(1+cosa)] =根号[2cos^(a/2)/(2sin^(a/2)]+根号[2sin^(a/2)/(2cos^(a/2)] =|cos(a/2)|/|sin(a/2)|+|sin(a/2)|/|cos(a/2)| =[cos^(a/2)+sin^(a/2)]/|sin(a/2)*cos(a/2)| =1/|(1/2)sina| =2/|sina| 。
sin(kt-a)cos(kt+a) -------------------------- sin[(k+1)t+a]cos[(k+1)t-a] (1/2)[sin(kt)+sin(-a)] =------------------------ (1/2)[sin(k+1)t+sina] sin(kt)-sina =---------------- sin(k+1)t+sina √[(1+cosa)/(1-cosa)]+√[(1-cosa)/(1+cosa)] =√[(1+cosa)²/(1-cos²a)]+√[(1-cosa)²/(1-cos²a)] =(1+cosa)/|sina|+(1-cosa)/|sina| =2/|sina|
答:1)2/(1*3)+2/(3*5)+2/(5*7)+……+2/99*101) !!! =(3-1)/(1*3)+(5-3)/(3*5)+(7-5)/(5*7)+...详情>>
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