Question

If \(f(x)=\sin x\), \(x\in[0,\pi/2]\) and \(g(x)=\cos x\), \(x\in[0,\pi/2]\), prove that \((f+g)\) is not one-one.

Hint:

Check the values of sin x plus cos x at the two endpoints of the interval.

Answer 1

Step 1: A Symmetry Based Route:
Instead of picking endpoints, use the identity connecting sin and cos of complementary angles.
We know $\sin(\pi/2-x)=\cos x$ and $\cos(\pi/2-x)=\sin x$ for any x.

Step 2: Build a General Identity for h:
Let $h(x)=\sin x+\cos x$. Replacing x by $\pi/2-x$ gives
\[ h(\pi/2-x)=\sin(\pi/2-x)+\cos(\pi/2-x)=\cos x+\sin x=h(x) \]
So $h(x)=h(\pi/2-x)$ holds for every x in the domain.

Step 3: Pick a Specific Pair of Distinct Points:
Choose $x_1=\pi/6$, so $x_2=\pi/2-\pi/6=\pi/3$. Both lie in $[0,\pi/2]$ and $x_1\ne x_2$.
Compute both values directly to confirm.
\[ h(\pi/6)=\sin(\pi/6)+\cos(\pi/6)=\dfrac12+\dfrac{\sqrt3}{2} \]
\[ h(\pi/3)=\sin(\pi/3)+\cos(\pi/3)=\dfrac{\sqrt3}{2}+\dfrac12 \]
These are equal, confirming the identity.

Final Answer:
Since $x_1=\pi/6$ and $x_2=\pi/3$ are distinct but give the same value of h, (f+g) is not one-one, hence proved. \[ \boxed{h(\pi/6)=h(\pi/3),\ \text{so } f+g \text{ is not one-one}} \]