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

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.

\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}}$$.