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Prove the following identities: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{2x}\text{ }+\text{ }\mathbf{3}\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }=\text{ }\mathbf{4}\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x})\]
Prove the following identities: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{2x}\text{ }+\text{ }\mathbf{3}\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }=\text{ }\mathbf{4}\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x})\]
CBSE | Class 11 | Exercise 9.1 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove the following identities: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{2x}\text{ }+\text{ }\mathbf{3}\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }=\text{ }\mathbf{4}\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x})\]

Take RHS:

\[4\text{ }(co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x)\]

By expanding we get,

\[4\text{ }(co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x)\text{ }=\text{ }4\text{ }[{{(co{{s}^{2}}~x)}^{3}}~\text{ }{{(si{{n}^{2}}~x)}^{3}}]\]

\[=\text{ }4\text{ }(co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x)\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\]

From Formula, \[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\]

So,

\[=\text{ }4\text{ }cos\text{ }2x\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x\text{ }\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[{{(co{{s}^{2}}~x)}^{2}}~+\text{ }{{(si{{n}^{2}}~x)}^{2}}~+\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x\text{ }\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x]\]

We know that, \[\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2ab\text{ }=\text{ }{{\left( a\text{ }+\text{ }b \right)}^{2}}\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[{{\left( 1 \right)}^{2}}~\text{ }1/4\text{ }(4\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)]\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[{{\left( 1 \right)}^{2}}~\text{ }1/4\text{ }{{\left( 2\text{ }cos\text{ }x\text{ }sin\text{ }x \right)}^{2}}]\]

From formula, \[sin\text{ }2x\text{ }=\text{ }2sin\text{ }x\text{ }cos\text{ }x\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[({{1}^{2}})\text{ }\text{ }1/4\text{ }{{\left( sin\text{ }2x \right)}^{2}}]\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }(1\text{ }\text{ }1/4\text{ }si{{n}^{2}}~2x)\]

From formula, \[\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\text{ }\text{ }co{{s}^{2}}~x\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[1\text{ }\text{ }1/4\text{ }(1\text{ }\text{ }co{{s}^{2}}~2x)]\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[1\text{ }\text{ }1/4\text{ }+\text{ }1/4\text{ }co{{s}^{2}}~2x]\]

\[=\text{ }4\text{ }cos\text{ }2x\text{ }[3/4\text{ }+\text{ }1/4\text{ }co{{s}^{2}}~2x]\]

\[=\text{ }4\text{ }(3/4\text{ }cos\text{ }2x\text{ }+\text{ }1/4\text{ }co{{s}^{3}}~2x)\]

\[=\text{ }3\text{ }cos\text{ }2x\text{ }+\text{ }co{{s}^{3}}~2x\]

\[=\text{ }co{{s}^{3}}~2x\text{ }+\text{ }3\text{ }cos\text{ }2x\]

= LHS

Hence proved.

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