Ifn=1,2,3....,then cosαcos2αcos3α·····cos2n-1α is equal to.
sin2nα2nsinα
sin2nα2nsin2n-1α
sin4n-1α4n-1sin4n-1α
Explanation for correct option
Given trigonometric expression is cosαcos2αcos3α·····cos2n-1α
∴cosαcos2αcos4α·····cos2n-1α=12sinα2sinαcosαcos2αcos4α·····cos2n-1α=12sinαsin2αcos2αcos4α·····cos2n-1α∵2sinθcosθ=sin2θ=122sinα2sin2αcos2αcos4α·····cos2n-1α=122sinαsin4αcos4α·····cos2n-1α∵2sinθcosθ=sin2θ
Proceeding in this way we get sin2nα2nsinα
Hence, option (D) i.e. sin2nα2nsinα is correct
If n is odd, then 1+3+5+7+.....+ to n terms is equal to
(a) (n2+1)
(b) (n2−1)
(c) n2
(d) (2n2+1)