SIN x=\cos x\RIGHTARROW x=\pi /4\] If \[y=\sin x\Rightarrow {{\LEFT( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{1}{\sqrt{2}}\] If \[y=\cos x\Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{-1}{\sqrt{2}}\] \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}=2\sqrt{2}\Rightarrow \,\theta ={{\tan }^{-1}}(2\sqrt{2})\].

"> SIN x=\cos x\RIGHTARROW x=\pi /4\] If \[y=\sin x\Rightarrow {{\LEFT( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{1}{\sqrt{2}}\] If \[y=\cos x\Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{-1}{\sqrt{2}}\] \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}=2\sqrt{2}\Rightarrow \,\theta ={{\tan }^{-1}}(2\sqrt{2})\].

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The angle between the curves \[y=\sin x\] and \[y=\cos x\] is [EAMCET 2003]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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If \[\SIN x=\cos x\RIGHTARROW x=\pi /4\] If \[y=\sin x\Rightarrow {{\LEFT( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{1}{\sqrt{2}}\] If \[y=\cos x\Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{-1}{\sqrt{2}}\] \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}=2\sqrt{2}\Rightarrow \,\theta ={{\tan }^{-1}}(2\sqrt{2})\].

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Posted on 24 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). 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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