# How to prove the value of cos a + sin a is greater than 1 ?

Dec 26, 2016

For $2 k \pi < a < \frac{\pi}{2} + 2 k \pi$ for $k = 0 , 1 , 2 , \cdots$

#### Explanation:

If $\cos \left(a\right) + \sin \left(a\right) > 1$ then ${\left(\cos \left(a\right) + \sin \left(a\right)\right)}^{2} = 1 + 2 \cos \left(a\right) \sin \left(a\right) > 1$ or
$\cos \left(a\right) \sin \left(a\right) > 0$
Concluding it is sufficient $\cos \left(a\right) \sin \left(a\right) > 0$ to verify
$\cos \left(a\right) + \sin \left(a\right) > 1$
or
$2 k \pi < a < \frac{\pi}{2} + 2 k \pi$ for $k = 0 , 1 , 2 , \cdots$

Attached a plot showing

$\cos \left(a\right) + \sin \left(a\right)$ in blue
$\max \left(\cos \left(a\right) \sin \left(a\right) , 0\right)$ in red and
the constant $1$ in green.

Dec 26, 2016

$\sin a + \cos a \in \left[- \sqrt{2} , \sqrt{2}\right]$.

#### Explanation:

For $a \in \left(- \infty , \infty\right)$,

$\cos a + \sin a = \sqrt{2} \left(\frac{1}{\sqrt{2}} \cos a + \frac{1}{\sqrt{2}} \sin a\right)$

=sqrt2(sin (pi/4)cos a+cos(pi/4)sin a

$= \sqrt{2} \sin \left(a + \frac{\pi}{4}\right) \in \left[- \sqrt{2} , \sqrt{2}\right]$

As a is not stated to be in an interval, I can assign $a = - \frac{\pi}{4}$ to get

0 for sin a + cos a.