# Is f(x)=sinx concave or convex at x=pi/5?

Aug 6, 2016

$f \left(x\right) = \sin x$ is concave at $\frac{\pi}{5}$

#### Explanation:

$f \left(x\right)$ is concave at ${x}_{0}$ if $f \left({x}_{0}\right) > \frac{1}{2} \left(f \left({x}_{0} - h\right) + f \left({x}_{0} + h\right)\right)$ and

$f \left(x\right)$ is convex at ${x}_{0}$ if $f \left({x}_{0}\right) < \frac{1}{2} \left(f \left({x}_{0} - h\right) + f \left({x}_{0} + h\right)\right)$,

where $\left({x}_{0} - h\right)$ and $\left({x}_{0} + h\right)$ are two values around ${x}_{0}$

As $\frac{\pi}{5} = {36}^{o}$, let us calculate $\sin x$ at $\left\{x = {35}^{o} , {36}^{o} , {37}^{o}\right\}$

$\sin {35}^{o} = 0.5735764$

$\sin {36}^{o} = 0.5877853$

$\sin {37}^{o} = 0.6018150$

Now as $\frac{1}{2} \left(\sin {35}^{\oplus} \sin {37}^{o}\right) = \frac{1}{2} \left(0.5735764 + 0.6018150\right)$

= $\frac{1}{2} \times 1.1753914 = 0.5876957 < \sin {36}^{o}$

$f \left(x\right) = \sin x$ is concave at $\frac{\pi}{5}$

Further if $f \left(x\right)$ can be differentiated twice at $x = {x}_{0}$ and $f ' ' \left({x}_{0}\right) < 0$, it is concave and if $f ' ' \left({x}_{0}\right) > 0$, it is convex.

As $f \left(x\right) = \sin x$, $f ' \left(x\right) = \cos x$ and $f ' ' \left(x\right) = - \sin x$ and

$f ' ' \left({36}^{o}\right) = - \sin {36}^{o} = - 0.5877853$ and hence $f \left(x\right) = \sin x$ is concave at $\frac{\pi}{5}$.