Question

Find the equation of the tangents to the curve \[y = \cos \left( {x + y} \right){\text{, }} - 2\pi \leqslant x \leqslant 2\pi \] that are parallel to the line \[x + 2y = 0\].

Answer 1

Hint: The tangent of an equation is the straight line which touches the curve at only one point. The equation of tangent passing through a point \[\left( {{x_0},{y_0}} \right)\] having slope of tangent \[m\] is given as \[y - {y_0} = m\left( {x - {x_0}} \right)\]. The slope of the tangent of two parallel lines is always equal. The slope of tangent of a curve is also equal to \[\dfrac{{dy}}{{dx}}\] at\[\left( {{x_0},{y_0}} \right)\]. The general equation of the line is \[y = mx + c\] where \[m\] the slope of tangent is.

Complete step by step answer:
The curve is \[y = \cos \left( {x + y} \right).............(i)\]
We have to find the equation of tangents to this curve which is parallel to the given line\[x + 2y = 0\]. Rearranging the equation of straight line in the general equation of line in order to get the slope of tangent.
\[x + 2y = 0 \\
\Rightarrow 2y = - x \\ \]
On dividing by \[2\] we get \[y = \dfrac{{ - 1}}{2}x + 0\]. On comparing with the general equation of line we get the slope of tangent \[m = \dfrac{{ - 1}}{2}\]. Now the slope of the tangent of a curve is also given as\[\dfrac{{dy}}{{dx}}\]. So on differentiating equation (i) we get,
\[\dfrac{{d\left( y \right)}}{{dx}} = \dfrac{{d\left\{ {\cos \left( {x + y} \right)} \right\}}}{{dx}}\]
On operating by chain rule of differentiation
\[\dfrac{{d\left( y \right)}}{{dx}} = \dfrac{{d\left\{ {\cos \left( {x + y} \right)} \right\}}}{{d\left( {x + y} \right)}} \times \dfrac{{d\left( {x + y} \right)}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin \left( {x + y} \right) \times \left( {1 + \dfrac{{dy}}{{dx}}} \right)...............(ii)\]
As the given straight line is parallel to the tangent of the curve so the slope of tangent will be equal So,
\[\dfrac{{dy}}{{dx}} = m \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{2} \\ \]
Substituting this in equation (ii) we get,
 \[\dfrac{{ - 1}}{2} = - \sin \left( {x + y} \right) \times \left( {1 - \dfrac{1}{2}} \right) \\
\Rightarrow \sin \left( {x + y} \right) = 1....................(iii) \\ \]
Taking inverse sine function in the equation we have,
\[x + y = {\sin ^{ - 1}}1 \\
\Rightarrow x + y = \dfrac{\pi }{2} \\ \]
And hence from equation (i) we get,
\[y = \cos \left( {x + y} \right) \\
\Rightarrow y = \cos \dfrac{\pi }{2} \\
\Rightarrow y= 0 \\ \]
For the coordinate of intersection point of curve and the tangent substituting\[y = 0\] in equation (iii) we get,
\[\sin \left( {x + y} \right) = 1 \\
\Rightarrow \sin x = 1 \\ \]
For the given interval
\[\sin x = 1 \\
\Rightarrow x = \dfrac{\pi }{2},\dfrac{{ - 3\pi }}{2} \\ \]
Hence the two points on the curve through which tangent could be drawn parallel to the given straight line are: \[\left( {\dfrac{\pi }{2},0} \right)\] and \[\left( {\dfrac{{ - 3\pi }}{2},0} \right)\]

For point \[\left( {{x_0},{y_0}} \right)\] the equation of tangent is given as \[y - {y_0} = m\left( {x - {x_0}} \right)\] where \[m\] is the slope of the tangent.Here slope of the tangent \[m = \dfrac{{ - 1}}{2}\]. Hence the equation of tangent through \[\left( {\dfrac{\pi }{2},0} \right)\] is
\[y - 0 = \dfrac{{ - 1}}{2}\left( {x - \dfrac{\pi }{2}} \right) \\
\Rightarrow y = \dfrac{{ - x}}{2} + \dfrac{\pi }{4} \\ \]
On taking L.C.M.
\[y = \dfrac{{ - 2x + \pi }}{4} \\
\Rightarrow 4y = - 2x + \pi \\ \]
Hence we get the equation of tangent through \[\left( {\dfrac{\pi }{2},0} \right)\] as
\[4y + 2x - \pi = 0\]
Again equation of tangent through \[\left( {\dfrac{{ - 3\pi }}{2},0} \right)\] is
\[y - 0 = \dfrac{{ - 1}}{2}\left( {x - \dfrac{{ - 3\pi }}{2}} \right) \\
\Rightarrow y = \dfrac{{ - x}}{2} - \dfrac{{3\pi }}{4} \\ \]
On taking L.C.M.
\[y = \dfrac{{ - 2x - 3\pi }}{4} \\
\therefore 4y = - 2x - 3\pi \]
Hence we get the equation of tangent through \[\left( {\dfrac{{ - 3\pi }}{2},0} \right)\] as
\[4y + 2x + 3\pi = 0\]

Therefore the equation of the tangents to the curve \[y = \cos \left( {x + y} \right){\text{, }} - 2\pi \leqslant x \leqslant 2\pi \] that are parallel to the line \[x + 2y = 0\] are \[4y + 2x - \pi = 0\] and \[4y + 2x + 3\pi = 0\].

Note:Slope of tangent is that tangent of angle made by the line in the positive direction of \[x - axis\]. The curve of cos function is symmetric to both \[x - axis\] and \[y - axis\]. Here in the given interval it crosses the \[x - axis\] twice so it has two equations of tangents. Similar is the situation for sine functions as well. The formulas for slope of tangents of a line are \[m = \tan \theta ,\,and\,\,m = \dfrac{{dy}}{{dx}}\].