FRAC{1}{x}=2\cos \THETA \], Now \[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3x\frac{1}{x}\left( x+\frac{1}{x} \right)\] = \[{{(2\cos \theta )}^{3}}-3(2\cos \theta )=8{{\cos }^{3}}\theta -6\cos \theta \] = \[2(4{{\cos }^{3}}\theta -3\cos \theta )=2\cos 3\theta \]. Trick: Put \[x=1\] \[\RIGHTARROW \] \[\theta ={{0}^{{}^\CIRC }}\]. Then\[{{x}^{3}}+\frac{1}{{{x}^{3}}}=2=2\cos 3\theta \].

"> FRAC{1}{x}=2\cos \THETA \], Now \[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3x\frac{1}{x}\left( x+\frac{1}{x} \right)\] = \[{{(2\cos \theta )}^{3}}-3(2\cos \theta )=8{{\cos }^{3}}\theta -6\cos \theta \] = \[2(4{{\cos }^{3}}\theta -3\cos \theta )=2\cos 3\theta \]. Trick: Put \[x=1\] \[\RIGHTARROW \] \[\theta ={{0}^{{}^\CIRC }}\]. Then\[{{x}^{3}}+\frac{1}{{{x}^{3}}}=2=2\cos 3\theta \].

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If \[x+\frac{1}{x}=2\,\cos \theta ,\] then \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=\] [MP PET 2004]

Mathematics Trigonometric Identities in Mathematics . 10 months ago

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We have\[x+\FRAC{1}{x}=2\cos \THETA \], Now \[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3x\frac{1}{x}\left( x+\frac{1}{x} \right)\] = \[{{(2\cos \theta )}^{3}}-3(2\cos \theta )=8{{\cos }^{3}}\theta -6\cos \theta \] = \[2(4{{\cos }^{3}}\theta -3\cos \theta )=2\cos 3\theta \]. Trick: Put \[x=1\] \[\RIGHTARROW \] \[\theta ={{0}^{{}^\CIRC }}\]. Then\[{{x}^{3}}+\frac{1}{{{x}^{3}}}=2=2\cos 3\theta \].

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Posted on 24 Aug 2024, this text provides information on Mathematics related to Trigonometric Identities in Mathematics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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