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Prove That: ( Cos α + Cos β 2 ) + ( Sin α + Sin β ) 2 = 4 Cos 2 ( α − β 2 ) - Mathematics

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Question

Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]

 
Numerical

Solution

\[LHS = \left( cos \alpha + cos\beta \right)^2 + \left( sin\alpha + sin\beta \right)^2 \]

\[ = \cos^2 \alpha + \cos^2 \beta + 2cos\alpha cos\beta + \sin^2 \alpha + \sin^2 \beta + 2sin\alpha sin\beta\]

\[ = ( \cos^2 \alpha + \sin^2 \alpha) + ( \cos^2 \beta + \sin^2 \beta) + 2\left( cos\alpha cos\beta + sin\alpha sin\beta \right)\]

\[ = 1 + 1 + 2\cos(\alpha - \beta)\]

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

\[ = 2\left\{ 2 \cos^2 \left( \frac{\alpha - \beta}{2} \right) \right\}\]

\[ = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right) = RHS\]

\[\text{ Hence proved } .\]

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 28]

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RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 11 | Page 28

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