To show that d/dz(sin z) = cos z
Solution:\\Method ~1: \\To Prove:\frac{d}{dz}(sin ~z)=cos ~z ~,where~z=x+iy \\We~know~that~ f(x+iy)=u(x,y)+iv(x,y) \\Let~f(z)=sin~z \\~~~~~~~~~~~~~~~~=sin ~(x+iy) \\~~~~~~~~~~~~~~~~=sin~x~cos(iy)~+~cos ~x~sin(iy)......................(I) \\Now ~we~ will ~find~ cos(iy)~and ~sin( iy ) \\cos(iy)=\frac{e^{i(iy)+e^{-i(iy)}}}{2}=\frac{e^{-y}+e^y}{2}=\frac{e^y+e^{-y}}{2}=cos(hy) \\ [since~i.i=i^2=-1~ and~-i.i=-i^2=-(-1)=1~] \\sin(iy)=\frac{e^{i(iy)-e^{-i(iy)}}}{2i}=\frac{e^{-y}-e^y}{2i}=i ~sin (hy) \\Now~ we ~put ~ values ~of ~cos(iy)~and ~sin(iy)~in ~equation ~(I), \\\therefore f(z)=sin~x~( cos(hy))~+~cos ~x~(i~sin(hy))\\~~~~~~~~~~~~=sin~x~( cos(hy))~+i~(cos ~x~ sin(hy)) \\Now~\frac{d}{dz}(sin~z)=\frac{d}{dz}[sin~x~( cos(hy))~+i~cos ~x~(sin(hy))] \\~~~~~~~~~~~~~~~~~~~~~~~~~~~=cos~x(cos(hy))+i~sin~~x(sin(hy)) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~=cos~x(cos(hy))+sin~x(i~sin(hy)) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~=cos~x(cos(iy))+sin~x(sin(iy)) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~=cos~x(cos(iy))-sin~x(sin(iy)) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~=cos(x+iy) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~=cos~z \\Method ~2: \\\frac{d}{dx}(sin ~z)=\frac{d}{dz}[\frac{e^{iz}-e^{-iz}}{2i}]=\frac{ie^{iz}+ie^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}{2}=cos~z
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