Section 1.4: Composition of Functions

Solution
Since for any two distinct elements \(x_1\) and \(x_2\) in \(\left[0, \frac{\pi}{2}\right]\),
\(\sin x_1 \neq \sin x_2\) and \(\cos x_1 \neq \cos x_2\), both \(f\) and \(g\) must be one-one.
But
\[ (f + g)(0) = \sin 0 + \cos 0 = 1 \]
and
\[ (f + g)\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 \]
Therefore, \(f + g\) is not one-one.