IIT JEE , AIEEE Forum - Mathematics , Trignometry

sneha_91

cos A+cos 2A+cos 3A

sneha_91

SOME ONE HELP ....I WANT TO KNOW THE SUM OF AN A.P IN TRIG

catch_arnnie

the formula for finding the sum of such series is as follows::

sum(S) = [ cos [ { (1st angle) + (last angle) }/2] sin ( n(difference)/2 ) ]/[sin {(difference)/2} ] where n is no. of terms. (difference) is common difference b/w angles.

i hope this helps..

iitkgp_bipin

Formula :

cosA+cos(A+d)+cos(A+2d)+.........+cos{A+(n-1)d} = cos[A+{(n-1)d/2}].sin(nd/2) / sin(d/2)

In the question given above put n=3 and d=A to get the answer.

cosA+cos2A+cos3A = cos2A.sin(3A/2) / sin(A/2)

waterdemon

I would like to ask BIPIN that how did u get the RHS in the above question .

nick

hiya dude,
the rhs above is through a derivation of complex root,
it is cos[A+{(n-1)d/2}].sin(nd/2) / sin(d/2) when angles of cos are in AP
and sin [A+{(n-1)d/2}].sin(nd/2) / sin(d/2) when angles of sin are in AP

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