`cos x+cos 2x+...+cos nx = sin((nx)/2) cosec(x/2)cos (((n+1)x)/2)`

cos x + cos2x + ... + cos nx = sin ((nx) / (2)) cos ec ((x) / (2)) cos (((n + 1) x) / (2))

(1 + sin 2x + cos 2x )/( cos x + sin x ) = 2 cos x

Solve cos x + cos 2x+...+ cos (nx) =n, n in N .

Solve cos x + cos 2x+...+ cos (nx) =n, n in N .

Solve cos x + cos 2x+...+ cos (nx) =n, n in N .

If x is not an integral multiple of 2 pi , use Mathematical Induction to prove that : cos x+cos 2x+.............+cos nx= cos frac (n+1)(2) x sin frac (nx)(2) cosec frac (x)(2) .

f (x) = (cos x) ( cos 2x ) …( cos nx) implies f' (x) + sum ^(n) (r tan rx) f (x) =

If maximum and minimum values of the determinant |{:(1 + cos^(2)x , sin^(2) x, cos 2x),(cos^(2) x , 1 + sin^(2)x, cos 2x),(cos^(2) x , sin^(2) x , 1 + cos 2 x):}| are alpha and beta then

If y =(cos x -sin x )/( cos x + si nx ) ,the n(dy)/(dx) =