#f(x)# is concave at #x_0# if #f(x_0)>1/2(f(x_0-h)+f(x_0+h))# and

#f(x)# is convex at #x_0# if #f(x_0)<1/2(f(x_0-h)+f(x_0+h))#,

where #(x_0-h)# and #(x_0+h)# are two values around #x_0#

As #pi/5=36^o#, let us calculate #sinx# at #{x=35^o,36^o,37^o}#

#sin35^o=0.5735764#

#sin36^o=0.5877853#

#sin37^o=0.6018150#

Now as #1/2(sin35^o+sin37^o)=1/2(0.5735764+0.6018150)#

= #1/2xx1.1753914=0.5876957< sin36^o#

#f(x)=sinx# is concave at #pi/5#

Further if #f(x)# can be differentiated twice at #x=x_0# 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#.