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Prove the following identities: \[\mathbf{cos}\text{ }\mathbf{4x}\text{ }=\text{ }\mathbf{1}\text{ }\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }+\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\]
Prove the following identities: \[\mathbf{cos}\text{ }\mathbf{4x}\text{ }=\text{ }\mathbf{1}\text{ }\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }+\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\]
CBSE | Class 11 | Exercise 9.1 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove the following identities: \[\mathbf{cos}\text{ }\mathbf{4x}\text{ }=\text{ }\mathbf{1}\text{ }\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }+\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\]

Take LHS:

\[cos\text{ }4x\]

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

So,

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

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

\[=\text{ }2[{{(2\text{ }co{{s}^{2}}~2x)}^{~2}}~+\text{ }{{1}^{2}}~\text{ }2\times 2\text{ }co{{s}^{2}}~x]\text{ }\text{ }1\]

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

\[=\text{ }8\text{ }co{{s}^{4}}~2x\text{ }+\text{ }2\text{ }\text{ }8\text{ }co{{s}^{2}}~x\text{ }\text{ }1\]

\[=\text{ }8\text{ }co{{s}^{4}}~2x\text{ }+\text{ }1\text{ }\text{ }8\text{ }co{{s}^{2}}~x\]

= RHS

Hence Proved.

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