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Prove the following identities: \[\mathbf{2}(\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x})\text{ }\text{ }\mathbf{3}(\mathbf{si}{{\mathbf{n}}^{\mathbf{4}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x})\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]
Prove the following identities: \[\mathbf{2}(\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x})\text{ }\text{ }\mathbf{3}(\mathbf{si}{{\mathbf{n}}^{\mathbf{4}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x})\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]
CBSE | Class 11 | Exercise 9.1 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove the following identities: \[\mathbf{2}(\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x})\text{ }\text{ }\mathbf{3}(\mathbf{si}{{\mathbf{n}}^{\mathbf{4}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x})\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]

Take LHS:

\[2(si{{n}^{6}}~x\text{ }+\text{ }co{{s}^{6}}~x)\text{ }\text{ }3(si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{4}}~x)\text{ }+\text{ }1\]

From formula, \[{{\left( a\text{ }+\text{ }b \right)}^{2}}~=\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2ab\]

\[{{a}^{3}}~+\text{ }{{b}^{3}}~=\text{ }\left( a\text{ }+\text{ }b \right)\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }ab)\]

So,

\[2(si{{n}^{6}}~x\text{ }+\text{ }co{{s}^{6}}~x)\text{ }\text{ }3(si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{4}}~x)\text{ }+\text{ }1\text{ }=~2\{{{(si{{n}^{2}}~x)}^{~3}}~+\text{ }{{(co{{s}^{2}}~x)}^{~3}}\}\text{ }\text{ }3\{{{(si{{n}^{2}}~x)}^{~2}}~+\text{ }{{(co{{s}^{2}}~x)}^{~2}}\}\text{ }+\text{ }1\]

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

\[=~2\{\left( 1 \right)\text{ }(si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{4}}~x\text{ }+\text{ }2\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\text{ }\text{ }3\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\}\text{ }\text{ }3\{{{(si{{n}^{2}}~x\text{ }+\text{ }co{{s}^{2}}~x)}^{~2}}~\text{ }2si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\}\text{ }+\text{ }1\]

We know, \[si{{n}^{2}}~x\text{ }+\text{ }co{{s}^{2}}~x\text{ }=\text{ }1\]

\[=~2\{{{(si{{n}^{2}}~x\text{ }+\text{ }co{{s}^{2}}~x)}^{~2}}~\text{ }3\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\}\text{ }\text{ }3\{{{\left( 1 \right)}^{2}}~\text{ }2si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\}\text{ }+\text{ }1\]

\[=~2\{{{\left( 1 \right)}^{2}}~\text{ }3\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\}\text{ }\text{ }3(1\text{ }\text{ }2si{{n}^{2}}~x\text{ }co{{s}^{2}}~x)\text{ }+\text{ }1\]

\[=~2(1\text{ }\text{ }3\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x)\text{ }\text{ }3\text{ }+\text{ }6\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\text{ }+\text{ }1\]

\[=~2\text{ }\text{ }6\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\text{ }\text{ }2\text{ }+\text{ }6\text{ }si{{n}^{2}}~x\text{ }co{{s}^{2}}~x\]

\[=\text{ }0\]

= RHS

Hence proved.

i

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