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). \[{\tan ^{ - 1}}\left( {5\sqrt 2 } \right)\]

VerifiedVerified
146.7k+ views
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.

Complete step-by-step answer:
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.