Question

Show that \(\sin x < x\) if \(0 < x < 2\pi \).

Short answer

It is proved that \(\sin x < x\) if \(0 < x < 2\pi \).

Step-by-step solution

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

Proof

Let the function \(f\left( x \right) = \sin x\). It is known that the function\(f\left( x \right) = \sin x\) is continuous and differentiable on \(\mathbb{R}\).

Suppose there is a number \(a\) such that \(0 < a < 2\pi \). So, \(f\) will be continuous on \(\left( {0,a} \right)\) and differentiable on \(\left( {0,a} \right)\). Therefore, by Mean Value Theorem, there exists a number \(c \in \left( {0,a} \right)\) such that \(f'\left( c \right) = \frac{{f\left( a \right) - f\left( 0 \right)}}{{a - 0}}\).

Solve the above equation as follows:

\(\begin{aligned}{c}\cos c &= \frac{{\sin a - \sin 0}}{{a - 0}}\\\cos c &= \frac{{\sin a}}{a}\\a \cdot \cos c &= \sin a\end{aligned}\)

It is known that \(\cos c < 1\) for all \(c \in \left( {0,2\pi } \right)\).

Therefore, \(a \cdot 1 > \sin a \Rightarrow a > \sin a\).

Since, \(a\) is an arbitrary number in \(\left( {0,2\pi } \right)\), therefore, \(\sin x < x\) for all \(x \in \left( {0,2\pi } \right)\).

Hence, it proved that \(\sin x < x\).