`cos(n+1)alphacos(n-1)alpha+sin(n+1)alpha sin(n-1)alpha=`

cos (n + 1) alpha cos (n-1) alpha + sin (n + 1) alpha sin (n-1) alpha =

cos(n+1) alpha * cos(n-1)alpha + sin (n+1)alpha * sin (n-1) alpha =

If a cos^(3)alpha+3a cos alpha sin^(2)alpha=ma sin^(3)alpha+3a cos^(2)alpha sin alpha=n, then (m+n)^((2)/(3))+(m-n)^((2)/(3))=

If alpha-beta=(2n+1)(pi)/2, n epsilon Z then [(cos^(2) alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)][(cos^(2) beta, cos beta sin beta),(cos beta sin beta, sin^(2) beta)]=

If a cos^(3)alpha+3a cos alpha*sin alpha=m and a sin^(3)alpha+3a cos^(3)alpha sin^(3)alpha=n then (m+n)^((2)/(3))+(m-n)^((2)/(3))

If a cos^(3)alpha+3a cos alpha sin^(2)alpha=m and a sin^(2)alpha+3a cos^(2)alpha sin alpha=n then (m+n)^((2)/(3))+(m-n)^((2)/(3))=

( sin(n+1) alpha - sin (n-1)alpha )/( cos (n+1) alpha + 2 cos n alpha + cos (n-1) alpha )=

If A_(alpha)=[[cos alpha, sin alpha],[ -sin alpha , cos alpha]] , then prove that i) A_(alpha ). A_(beta)=A_(alpha + beta) ii) (A_alpha)^n=[[cos n alpha, sin n alpha],[ -sin n alpha, cos n alpha]] for every positive integer n .

If n-1,2,3,dot,t h e ncosalpha"cos"2alpha"cos"4alphacos2^(n-1)alphai se q u a lto (sin2nalpha)/(2nsinalpha) (b) (sin2^nalpha)/(2^n"sin"2^("n"-1)alpha) (c) (sin4^(n-1)alpha)/(4^(n-1)sinalpha) (d) (sin2^nalpha)/(2^n"sin"alpha)