Prove that $-\sqrt{2}\leq \sin\theta+\cos\theta\leq\sqrt{2}$ We have to do it without calculus or any complex inequality. Level of complexity is that we cannot even use the AM-GM inequality. 
So I tried, 
$$(\sin\theta-\cos\theta)^2\geq0$$
$$1-2\sin\theta\cos\theta\geq0$$
$$\frac12\geq\sin\theta\cos\theta$$
Reverting back to the previous step, $$(\sin\theta+\cos\theta)^2\geq4\sin\theta\cos\theta$$
I am stuck here, please help. 
Edit: Sorry, but we can only use knowledge upto class 10th. Which includes,  $\sin^2\theta+\cos^2\theta=1$ etc. 
$\sin(90^{\circ}-\theta)=\cos\theta$ etc.
 And basic trig ratios.
 A: Hint: We have $$\sin\theta+\cos\theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right).$$Clearly we have $$-1\leq\sin(\theta+\pi/4)\leq1.$$
Update: To answer the edited question (that is, not using the sum formula for $\sin(\theta+\pi/4)$, we have $$(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\cdot\underbrace{(\sin^2\theta+\cos^2\theta)}_1+2(\sin\theta)(\cos\theta)-2(\sin\theta)(\cos\theta)=2.$$ Since $x^2\geq0$ for all real $x$, we can subtract $(\sin\theta-\cos\theta)^2$ from both sides to obtain $$2-(\sin\theta-\cos\theta)^2\geq0.$$Equivalently, this can be written $$(\sin\theta-\cos\theta)^2\leq2.$$This just means $$|\sin\theta-\cos\theta|\leq\sqrt{2}.$$ By definition of absolute value, this says $$-\sqrt{2}\leq\sin\theta-\cos\theta\leq\sqrt{2}.$$
A: There are easier ways, but we can use AM-GM.
By the Triangle Inequality we have $|\sin\theta+\cos\theta|\le |\sin\theta|+|\cos\theta|$.
By AM-GM we have $|\sin\theta|+|\cos\theta|\le \frac{1}{\sqrt{2}}|\sin\theta||\cos\theta|$. This is equal to $\sqrt{2}|\sin(2\theta)|$, which is $\le \sqrt{2}$.
Remark: You were almost finished, for $4\sin\theta\cos\theta=2\sin(2\theta)$.  And $2\sin(2\theta)$ has absolute value $\le 2$.
A: Using $$\displaystyle (\sin \phi+\cos \phi)^2+(\sin \phi-\cos \phi)^2 = 2$$
Now $$\displaystyle (\sin \phi-\cos \phi)^2 = 2-(\sin \phi-\cos \phi)^2$$
Using $$\bf{Square\; Quantity\geq 0}$$
So $$\displaystyle (\sin \phi-\cos \phi)^2\geq0$$
So $$\displaystyle 2-\displaystyle (\sin \phi+\cos \phi)^2\geq 0$$
OR we get $$\displaystyle (\sin \phi+\cos \phi)^2\leq \left(\sqrt{2}\right)^2$$
So  we get $$\displaystyle-\sqrt{2} \leq (\sin \phi+\cos \phi)\leq \sqrt{2}$$
A: A (simple?) answer:
Since $\sin$ and $\cos$ are bounded, there is $r>0$ such that
$$\left|\sin\theta+\cos\theta\right|\leq r$$
Replacing $\theta$ by $-\theta$ gives
$$\left|\sin\theta-\cos\theta\right|\leq r$$
Multiplying together
$$\left|\sin^2\theta-\cos^2\theta\right|\leq r^2$$
But the LHS is bounded by $2$, so picking $r$ smallest possible, we get that $r^2\leq 2$. So $r\leq\sqrt{2}$.
A: You could use the R-Formula (which can be derived using trigonometric identities) to combine $\sin \theta + \cos \theta$ into
$$\sqrt{2}\sin \left(\theta + \frac{\pi}{4}\right)$$
and then use the fact that $-1 \le \sin x \le 1$. 
A: Hint: Write $$\cos(b)=\sin(\frac{\pi}{2}-b)$$
Now use the formula of $\sin(a)+\sin(b)$
A: 1st PROOF:  $\sin \pi/4 =\cos \pi/4 =1/\sqrt 2$.   Therefore $|\sin x + \cos x|=\sqrt 2 |\sin x \cos \pi/4 +\cos x \sin \pi/4|=\sqrt 2 |\sin (x+\pi/4)|\le \sqrt 2$.Equality iff $|\sin (x+\pi/4)|=1.$ E.g. $x= \pi/4.$...............2nd PROOF:  $(\sin x +\cos x)^2 \le (\sin x +\cos x)^2+(\sin x - \cos x)^2 =2$. Equality iff  $\sin x - \cos x =0$ and $|\sin x=1/\sqrt 2|.$                    E.g. $x=\pi/4.$
A: $$-\sqrt{a^2+b^2} \le a\sin \theta+b\cos \theta \le \sqrt{a^2+b^2}$$
Plug in $a=b=1 $ in above formula.
You get:
$$-\sqrt{2} \le \sin \theta+\cos \theta \le \sqrt{2}$$
How I got the formula
Consider $f(\theta)=a\sin \theta+b\cos \theta$
Now I will differentiate this function to get min/max.
$$f^{'}(\theta)=a\cos \theta-b\sin \theta=0$$(for min/max)
$$a\cos \theta=b\sin \theta$$
$$\frac{a}{b}=\tan \theta$$
When $\tan \theta=\frac{a}{b}$ then there is maxima and minima.
Now you can find that $\sin \theta=\frac{a}{\sqrt{a^2+b^2}}$ and $\cos \theta=\frac{b}{\sqrt{a^2+b^2}}$
Plug in these values in the formula and you'll get it.
