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Prove that Cos α + Cos ( α + β ) + Cos ( α + 2 β ) + . . . + Cos [ α + ( N − 1 ) β ] = Cos { α + ( N − 1 2 ) β } Sin ( N β 2 ) Sin ( β 2 ) for All N ∈ N - Mathematics

\[\text{ Prove that } \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all n } \in N .\]

 

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Solution

\[\text{ Let p } \left( n \right): \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \forall n \in  N . \] 

\[\text{ Step I: For }  n = 1, \]

\[LHS = \cos\left[ \alpha + \left( 1 - 1 \right)\beta \right] = \cos\alpha\]

\[RHS = \frac{\cos\left\{ \alpha + \left( \frac{1 - 1}{2} \right)\beta \right\}\sin\left( \frac{\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} = \cos\alpha\]

\[\text{ As, LHS = RHS } \]

\[\text{ So, it is true for n = 1 .}  \]

\[\text{ Step II: For n = k,}  \]

\[ \text{ Let } p\left( k \right): \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( k - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{k - 1}{2} \right)\beta \right\}\sin\left( \frac{k\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ be true  } \forall k \in N . \]

\[\text{ Step III: For n  } = k + 1, \]

\[LHS = \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( k - 1 \right)\beta \right] + \cos\left[ \alpha + \left( k + 1 - 1 \right)\beta \right]\]

\[ = \frac{\cos\left\{ \alpha + \left( \frac{k - 1}{2} \right)\beta \right\}\sin\left( \frac{k\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} + \cos\left( \alpha + k\beta \right)\]

\[ = \frac{\cos\left\{ \alpha + \left( \frac{k - 1}{2} \right)\beta \right\}\sin\left( \frac{k\beta}{2} \right) + \sin\left( \frac{\beta}{2} \right)\cos\left( \alpha + k\beta \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{\sin\left( \alpha + k\beta - \frac{\beta}{2} \right) - \sin\left( \alpha - \frac{\beta}{2} \right) + \sin\left( \alpha + k\beta + \frac{\beta}{2} \right) - \sin\left( \alpha + k\beta - \frac{\beta}{2} \right)}{2\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{- \sin\left( \alpha - \frac{\beta}{2} \right) + \sin\left( \alpha + k\beta + \frac{\beta}{2} \right)}{2\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{2\cos\left( \frac{2\alpha + k\beta}{2} \right)\sin\left( \frac{k\beta + \beta}{2} \right)}{2\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{\cos\left( \alpha + \frac{k\beta}{2} \right)\sin\left( \frac{\left( k + 1 \right)\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[RHS = \frac{\cos\left\{ \alpha + \left( \frac{k + 1 - 1}{2} \right)\beta \right\}\sin\left( \frac{\left( k + 1 \right)\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{\cos\left( \alpha + \frac{k\beta}{2} \right)\sin\left( \frac{\left( k + 1 \right)\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[As, LHS = RHS\]

\[\text{ So, it is also true for n = k + 1 .} \]

\[\text{ Hence,}  \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all } n \in N .\]

 

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 40 | Page 29
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