Consider a function $f: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $g(x) = \cos x$. Show that $f$ and $g$ are one-one,but $f + g$ is not one-one.

(N/A) For any two distinct elements $x_1, x_2 \in [0, \frac{\pi}{2}]$,we know that the sine function is strictly increasing on this interval,so $\sin x_1 \neq \sin x_2$. Thus,$f$ is one-one.
Similarly,the cosine function is strictly decreasing on $[0, \frac{\pi}{2}]$,so $\cos x_1 \neq \cos x_2$. Thus,$g$ is one-one.
Now,consider the function $h(x) = (f + g)(x) = \sin x + \cos x$.
We evaluate $h(0) = \sin 0 + \cos 0 = 0 + 1 = 1$.
We evaluate $h(\frac{\pi}{2}) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$.
Since $h(0) = h(\frac{\pi}{2})$ but $0 \neq \frac{\pi}{2}$,the function $f + g$ is not one-one.