<s>[INST] Find the highest and the lowest value of the function $y=1+\sin x \cos x$
Please dont use the formula $2\sin x \cos x=\frac{1}{2}\sin 2x$. We have't learnt it yet.
The key is to return it into an inequation for example $-1 \le \sin x \le 1$ 
Then multiply with something to give an inequality
For example find the highest and lowest value of $y=\frac{1}{2}\sin x$
$$-1 \le \sin x \le 1$$
$$-\frac{1}{2} \le\frac{1}{2}\sin x \le \frac{1}{2}$$
The highest value is $\frac{1}{2}$, the lowest is $-\frac{1}{2}$
Could you solve $y=1+\sin x\cos x$ in this way after you have done equivalent changes?
 [/INST] We want to maximize/minimize $(\sin x)(\cos x)$, or equivalently $2(\sin x)(\cos x)$. Note that 
$$2\sin x\cos x=(\sin^2 x+\cos^2 x)-(\sin x-\cos x)^2=1-(\sin x-\cos x)^2.$$
This attains a maximum  when $(\sin x-\cos x)^2$ is as small as possible, namely $0$. 
For the minimum, use the same idea, and
$$2\sin x\cos x=(\sin x+\cos x)^2 -(\sin^2 x+\cos^2 x).$$
</s>