sin((n+1)/(2))theta=sin((n-1)/(2))theta+sin theta

Delta_(r)=det[[2^(r-1),(1)/(r(r+1)),sin r thetax,y,z2^(n)-1,(n)/(n+1),((sin)(n+1)/(2)theta(sin)(n)/(2)theta)/((sin theta)/(2))]], then sum_(r=1)^(n)Delta_(r)

Prove that , sin theta + sin 2theta + ……+ sin n theta = ( (sin( n theta/(2)) sin ((n+1)/(2))theta))/(sin theta/2) , for all n in N .

D_ (k) = | 2 ^ (k-1) (1) / (k (k + 1)) sin k theta xyz2 ^ (n) -1 (n) / (n + 1) (sin ((n +) 1) / (2) theta (sin n) / (2) theta) / (sin theta) / 2, thensum_ (k = 1) (n) D_ (k)

The sum of the series sin theta+sin((n-4)/(n-2))theta+sin((n-6)/(n-2))theta+......*n terms is equal to

Solve the equation sin^(2)ntheta - sin^(2)(n-1)theta = sin^(2)theta

Solve the equation sin^(2)ntheta - sin^(2)(n-1)theta = sin^(2)theta

Prove that sin theta+sin3 theta+sin5 theta+....+sin(2n-1)theta=(sin^(2)n theta)/(sin theta)

sin^2 n theta- sin^2 (n-1)theta= sin^2 theta where n is constant and n != 0,1

sin^(2)n theta-sin^(2)(n-1)theta=sin^(2)theta where n is constant and n!=0,1

sin^2 n theta- sin^2 (n-1)theta= sin^2 theta where n is constant and n != 0,1