`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)

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

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

Prove that, sintheta+sin2theta+sin3theta+ . . .sinntheta=(sin""(ntheta)/(2)"sin"(n+1)/(2)theta)/("sin"(theta)/(2)) for all ninN .

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

sin4 theta can be written as (a) 4sin theta(1-2sin^(2)theta)sqrt(1-sin^(2)theta)(b)2sin theta cos theta sin^(2)theta(c)4sin theta-6sin^(3)theta(d) None of the above

Prove that sin^(2)theta+sin^(2)2 theta+sin^(2)3 theta+....+sin^(2)n theta=(n)/(2)-(sin n theta cos(n+1)theta)/(2sin theta)

Prove that sum_(r=1)^(n)((1)/(cos theta+cos(2r+1)theta))=(sin n theta)/(2sin theta*cos theta*cos(n+1)theta),(wherenin N)