# The angle between the curves $y = \sin x$ and $y = \cos x$ is A). ${\tan ^{ - 1}}\left( {2\sqrt 2 } \right)$B). ${\tan ^{ - 1}}\left( {3\sqrt 2 } \right)$C). ${\tan ^{ - 1}}\left( {3\sqrt 3 } \right)$D). ${\tan ^{ - 1}}\left( {5\sqrt 2 } \right)$

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Hint: We know that $y = \sin x = \cos x$ when $x = \dfrac{\pi }{4}$ use this to find the slopes of both the curves then usen the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ where $m_1$ and $m_2$ are slopes of given curves.

Let the two curves be ${C_1}$ and ${C_2}$ respectively
${C_1}:y = \sin x,{C _ 2}:y = \cos x$
Equation both the curves we are getting
$\begin{array}{l} \sin x = \cos x\\ \therefore x = \dfrac{\pi }{4} \end{array}$
Therefore curves intersect each other at the point P : $x = \dfrac{\pi }{4}$
Now, differentiating with respect to. x,
${C_1}$ gives : $\dfrac{{dy}}{{dx}} = \cos x$
${C_2}$ gives : $\dfrac{{dy}}{{dx}} = - \sin x$
Hence the slopes ${{m_1}\& {m_2}}$ of ${C_1}\& {C_2}$ at P: $x = \dfrac{\pi }{4}$ are
$\begin{array}{l} {m_1} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\\ {m_2} = - \sin \dfrac{\pi }{4} = - \dfrac{1}{{\sqrt 2 }} \end{array}$
If $\theta$ is the acute angle between them at P, then
$\begin{array}{l} \therefore \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\\ \Rightarrow \tan \theta = \left| {\dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}} \right) - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}{{1 + \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}} \right|\\ \Rightarrow \tan \theta = \dfrac{{\left| {2\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right|}}{{\left| {\dfrac{{\left( {2 - 1} \right)}}{2}} \right|}}\\ \Rightarrow \tan \theta = 2\sqrt 2 \\ \therefore \theta = {\tan ^{ - 1}}\left( {2\sqrt 2 } \right) \end{array}$
So it can be clearly seen that option a is the correct option here.

Note: One must know the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ in order to solve these types of questions also remember that if we differentiate any function we get the slope of the curve formed by that function.