`sum_(n=1)^(20)[sin((2n pi)/(21))-i cos((2n pi)/(21))]=`

sum_(m=1)^(4n+1)[sum_(K=1)^(m-1){sin((2k pi)/(m))-i cos((2k pi)/(m))}^(m)]

lim_ (n rarr oo) sum_ (k = 1) ^ (n) ((sin) (pi) / (2k) - (cos) (pi) / (2k) - (sin) ((pi) / (2) k + 2)) + (cos) (pi) / (2 (k + 2))) =

Value of cos((pi)/(2n+1))+cos((3 pi)/(2n+1))+cos((5 pi)/(2n+1))+ upto n terms equals

lim_(n rarr oo)n sin\ (2 pi)/(3n)*cos\ (2 pi)/(3n)

sum_(i=1)^(2n)sin^(-1)(x_(i))=n pi then the value of sum_(i=1)^(n)cos^(-1)x_(i)+sum_(i=1)^(n)tan^(-1)x_(i)=(A)(n pi)/(4)(B)((2)/(3))n pi(C)((5)/(4))n pi(D)2n pi

The value of cos(2 pi)/(21)+cos(4 pi)/(21)+cos(6 pi)/(21)+....+cos(20 pi)/(21) is equal to

Show that if n>2cos^(2)((pi)/(n))+cos^(2)((3 pi)/(n))+cos^(2)((5 pi)/(n))+...nterms=(n)/(2)

The value of lim_(n rarroo) sum_(r=1)^(n)(1)/(sin{((n+r)pi)/(4n)}).(pi)/(n) is equal to

Prove that sum_(k=1)^(n-1)(n-k)(cos(2k pi))/(n)=-(n)/(2) wheren >=3 is an integer

The general solution of the equation 8cos x cos2x cos4x=(sin6x)/(sin x) is x=((n pi)/(7))+((pi)/(21)),AA n in Zx=((2 pi)/(7))+((pi)/(14)),AA n in Zx=((n pi)/(7))+((pi)/(14)),AA n in Zx=(n pi)+((pi)/(14)),AA n in Z