Evaluate `cos alpha cos 2 alpha cos 3 alpha..........cos 999alpha`, where `alpha=(2pi)/(1999)`

cos alpha cos 2 alpha cos 4 alpha cos 8 alpha = (1)/(16), if alpha = 24 ^(@)

If alpha = (pi)/(13), Prove that cos alpha cos 2alpha cos 3alpha cos 4alpha cos 5alpha cos 6alpha = (1)/(64) .

If alpha = pi/3, prove that cos alpha*cos 2alpha*cos 3alpha*cos 4alpha*cos 5alpha*cos 6alpha = -1/16.

If alpha = pi/3 , prove that cos alpha*cos 2alpha*cos 3alpha*cos 4alpha*cos 5alpha*cos 6alpha = -1/16 .

If alpha = pi/3 , prove that cos alpha*cos 2alpha*cos 3alpha*cos 4alpha*cos 5alpha*cos 6alpha = -1/16 .

Evaluate |["cos" alpha, -"sin"alpha], ["sin"alpha, "cos"alpha]|

Find the value of sin^6 alpha+ cos^6 alpha+ 2sin^2 alpha cos^2 alpha .

Find the sum to n terms of the following series: cos alpha + cos (pi - alpha) + cos (2pi + alpha) + .......(0 < alpha < (pi)/2) .

(sin 5 alpha-sin3alpha)/(cos 5 alpha+2 cos 4 alpha+cos 3alpha)=

cos2 alpha-cos3alpha-cos4alpha+cos5alpha simplifies to :