# 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.

• Are you aware that any linear combination of a sine and cosine of period $2\pi$ may be written as $c\sin(x+\theta)$ for some $c$ and $\theta$? (If not, try to find what $c$ and $\theta$ are) – Milo Brandt Sep 12 '15 at 14:05
• In your proof, use the fact that $4\sin\theta\cos\theta=2\sin(2\theta)$. This has absolute value $\le 2$. – André Nicolas Sep 12 '15 at 14:18
• Thank you but, I cannot use this, because I am currently studying in class tenth and I cannot use this. – Aditya Agarwal Sep 12 '15 at 14:20
• LESS OR EQUAL not "less than" – DanielWainfleet Sep 12 '15 at 14:21
• Sorry didn't get you @user254665 – Aditya Agarwal Sep 12 '15 at 14:22

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 $$-\sqrt{2}\leq\sin\theta-\cos\theta\leq\sqrt{2}.$$

• Please see the edit. Sorry for the mishap. – Aditya Agarwal Sep 12 '15 at 14:21
• Was just going to accept the answer, but can you please tell a way without using absolute value? It is out of course for us. In exams I wont be able to write. Rest all I can. Thanks for that. – Aditya Agarwal Sep 13 '15 at 4:05

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 $\le 2$.

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 $$\displaystyle-\sqrt{2} \leq (\sin \phi+\cos \phi)\leq \sqrt{2}$$

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 $r\leq\sqrt{2}$.

• When you replace $\theta$ with $-\theta$, the sine term takes on a negative sign out front because it's an odd function. The cosine term stays positive because it's an even function. That is, $\sin(-\theta) = -\sin(\theta)$ and $\cos(-\theta) = \cos(\theta)$. – Xoque55 Sep 12 '15 at 15:12
• Sure, but we only care about the absolute value of the expression, so it doesn't matter. – Damian Reding Sep 12 '15 at 17:41

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 $-1 \le \sin x \le 1$.

Hint: Write $$\cos(b)=\sin(\frac{\pi}{2}-b)$$ Now use the formula of $\sin(a)+\sin(b)$

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. $x=\pi/4.$