Answer
$$\left( {\text{a}} \right)y = x,\,\,\,\,\,\left( {\text{b}} \right)y = - x + \pi ,\,\,\,\,\,\left( {\text{c}} \right)y = x - \frac{{\sqrt 2 \pi }}{8} + \frac{{\sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \sin x \cr
& f'\left( x \right) = \cos x \cr
& \cr
& \left( {\text{a}} \right){\text{ At }}x = 0 \cr
& \,\,\,\,f\left( 0 \right) = \sin \left( 0 \right) = 0 \cr
& {\text{We obtain the point }}\left( {0,0} \right) \cr
& {\text{The slope at }}x = 0{\text{ is given by}} \cr
& \,\,m = f'\left( 0 \right) = \cos \left( 0 \right) \cr
& \,\,m = 1 \cr
& {\text{The equation of the line tangent to the graph is}} \cr
& \,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,y - 0 = 1\left( {x - 0} \right) \cr
& \,\,y = x \cr
& \cr
& \left( {\text{b}} \right){\text{ At }}x = \pi \cr
& \,\,\,\,f\left( \pi \right) = \sin \left( \pi \right) = 0 \cr
& {\text{We obtain the point }}\left( {\pi ,0} \right) \cr
& {\text{The slope at }}x = 0{\text{ is given by}} \cr
& \,\,m = f'\left( \pi \right) = \cos \left( \pi \right) \cr
& \,\,m = - 1 \cr
& {\text{The equation of the line tangent to the graph is}} \cr
& \,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,y - 0 = - 1\left( {x - \pi } \right) \cr
& \,\,y = - x + \pi \cr
& \cr
& \left( {\text{c}} \right){\text{ At }}x = \frac{\pi }{4} \cr
& \,\,\,\,\,f\left( {\frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \cr
& {\text{We obtain the point }}\left( {\frac{\pi }{4},\frac{{\sqrt 2 }}{2}} \right) \cr
& {\text{The slope at }}x = 0{\text{ is given by}} \cr
& \,\,m = f'\left( {\frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) \cr
& \,\,m = \frac{{\sqrt 2 }}{2} \cr
& {\text{The equation of the line tangent to the graph is}} \cr
& \,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,y - \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}\left( {x - \frac{\pi }{4}} \right) \cr
& \,\,y = x - \frac{{\sqrt 2 \pi }}{8} + \frac{{\sqrt 2 }}{2} \cr} $$
