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Prove the following: sin [(n+1)A] . sin [(n+2)A] + cos [(n+1)A] . cos [(n+2)A] = cos A |
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Answer» L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A] = cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A] Let(n+2)Aa and(n+l)Ab …(i) ∴ L.H.S. = cos a. cos b + sin a. sin b = cos (a — b) = cos [(n + 2)A — (n + I )A] …[From (i)] cos[(n+2 – n – 1)A] = cos A = R.H.S. |
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