If #f(r)=cos2rtheta# ,by simplifying f(r+1)-f(r).How to use the result to find the sum of first n terms of the series #sin3theta + sin5theta + sin7theta +...#?

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Answer 1 · 2017-08-06T16:12:08

Given

#f(r)=cos2rtheta#

So

#f(r+1)-f(r)=cos2(r+1)theta-cos2rtheta#

#=>f(r+1)-f(r)=-2sin(2r+1)thetasintheta#

#=>f(r)-f(r+1)=2sin(2r+1)thetasintheta#

Now putting #r=1,2,3,4.....n# and adding we get

for #r=1 tof(1)-f(2)=2sin(3theta)sintheta#

for #r=2 tof(2)-f(3)=2sin(5theta)sintheta#

for #r=3 tof(3)-f(4)=2sin(7theta)sintheta#

#-------------#

for #r=n tof(n)-f(n+1)=2sin((2n+1)theta)sintheta#

Adding we get

#f(1)-f(n+1)=2sintheta(sin3theta+sin5theta+………"upto n terms")#

So

#(sin3theta+sin5theta+………"upto n terms")=(f(1)-f(n+1))/(2sintheta)#

#=(cos2theta-cos2(n+1)theta)/(2sintheta)#

#=(2sin(n+2)thetasinntheta)/(2sintheta)#

#=(sin(n+2)thetasinntheta)/sintheta#