Prove that: ^2(n+1)A- ^2nA= (2n+1)A

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Prove that: $\sin^2\text{(n+1)}\text{A}-\sin^2\text{nA}=\sin\text{(2n+1)}\text{A}\sin\text{A}$

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Answer

We have, $\text{L.H.S} =\sin^2\text{(n+1)}\text{A}-\sin^2\text{nA}$ $=\sin[(\text{n+1)}\text{A}+\text{nA}]\sin[(\text{n+1)}\text{A}-\text{nA}]$ $\Big[\because\sin^2\text{A}-\sin^2\text{B}=\sin\text{(A+ B)}\sin\text{(A-B)}\Big]$ $=\sin[\text{nA}+\text{A}+\text{nA}]\sin[\text{nA}+\text{A}-\text{nA}]$ $=\sin\text{(2nA+A)}\sin\text{(A)}$ $=\sin\text{(2n+1)}\text{A}\sin\text{A}$ $=\text{R.H.S}$ $$$\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.

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