f(x)f(x) is concave at x_0x0 if f(x_0)>1/2(f(x_0-h)+f(x_0+h))f(x0)>12(f(x0−h)+f(x0+h)) and
f(x)f(x) is convex at x_0x0 if f(x_0)<1/2(f(x_0-h)+f(x_0+h))f(x0)<12(f(x0−h)+f(x0+h)),
where (x_0-h)(x0−h) and (x_0+h)(x0+h) are two values around x_0x0
As pi/5=36^oπ5=36o, let us calculate sinxsinx at {x=35^o,36^o,37^o}{x=35o,36o,37o}
sin35^o=0.5735764sin35o=0.5735764
sin36^o=0.5877853sin36o=0.5877853
sin37^o=0.6018150sin37o=0.6018150
Now as 1/2(sin35^o+sin37^o)=1/2(0.5735764+0.6018150)12(sin35⊕sin37o)=12(0.5735764+0.6018150)
= 1/2xx1.1753914=0.5876957< sin36^o12×1.1753914=0.5876957<sin36o
f(x)=sinxf(x)=sinx is concave at pi/5π5
Further if f(x)f(x) can be differentiated twice at x=x_0x=x0 and f''(x_0)<0, it is concave and if f''(x_0)>0, it is convex.
As f(x)=sinx, f'(x)=cosx and f''(x)=-sinx and
f''(36^o)=-sin36^o=-0.5877853 and hence f(x)=sinx is concave at pi/5.
