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Prove the following identities: \[{{\left( \mathbf{cos}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~+\text{ }{{\left( \mathbf{sin}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~=\text{ }\mathbf{4}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\left( \mathbf{\alpha }\text{ }\text{ }\mathbf{\beta } \right)/\mathbf{2}\]
Prove the following identities: \[{{\left( \mathbf{cos}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~+\text{ }{{\left( \mathbf{sin}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~=\text{ }\mathbf{4}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\left( \mathbf{\alpha }\text{ }\text{ }\mathbf{\beta } \right)/\mathbf{2}\]
CBSE | Class 11 | Exercise 9.1 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove the following identities: \[{{\left( \mathbf{cos}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~+\text{ }{{\left( \mathbf{sin}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~=\text{ }\mathbf{4}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\left( \mathbf{\alpha }\text{ }\text{ }\mathbf{\beta } \right)/\mathbf{2}\]

 

Take LHS:

\[{{\left( cos\text{ }\alpha \text{ }+\text{ }cos\text{ }\beta  \right)}^{2}}~+\text{ }{{\left( sin\text{ }\alpha \text{ }+\text{ }sin\text{ }\beta  \right)}^{2}}\]

By expanding we get,

\[{{\left( cos\text{ }\alpha \text{ }+\text{ }cos\text{ }\beta  \right)}^{2}}~+\text{ }{{\left( sin\text{ }\alpha \text{ }+\text{ }sin\text{ }\beta  \right)}^{2}}~=\]

\[=\text{ }co{{s}^{2}}~\alpha \text{ }+\text{ }co{{s}^{2}}~\beta \text{ }+\text{ }2\text{ }cos\text{ }\alpha \text{ }cos\text{ }\beta \text{ }+\text{ }si{{n}^{2}}~\alpha \text{ }+\text{ }si{{n}^{2}}~\beta \text{ }+\text{ }2\text{ }sin\text{ }\alpha \text{ }sin\text{ }\beta \]

\[=\text{ }2\text{ }+\text{ }2\text{ }cos\text{ }\alpha \text{ }cos\text{ }\beta \text{ }+\text{ }2\text{ }sin\text{ }\alpha \text{ }sin\text{ }\beta \]

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

\[=\text{ }2\text{ }\left( 1\text{ }+\text{ }cos\text{ }\left( \alpha \text{ }\text{ }\beta  \right) \right)\text{ }\left[ since,\text{ }cos\text{ }\left( A\text{ }\text{ }B \right)\text{ }=\text{ }cos\text{ }A\text{ }cos\text{ }B\text{ }+\text{ }sin\text{ }A\text{ }sin\text{ }B \right]\]

\[=\text{ }2\text{ }(1\text{ }+\text{ }2\text{ }co{{s}^{2}}~\left( \alpha \text{ }\text{ }\beta  \right)/2\text{ }\text{ }1)\text{ }[since,\text{ }cos2x\text{ }=\text{ }2co{{s}^{2}}~x\text{ }\text{ }1]\]

\[=\text{ }2\text{ }(2\text{ }co{{s}^{2}}~\left( \alpha \text{ }\text{ }\beta  \right)/2)\]

\[=\text{ }4\text{ }co{{s}^{2}}~\left( \alpha \text{ }\text{ }\beta  \right)/2\]

= RHS

Hence Proved.

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