# Thread: Prove that sin(x)+cos(x) is equal or larger than 1

1. We could use Lagrange Multipliers...consider the objective function:

$\displaystyle f(x,y)=\sin(x)-\sin(y)$

Subject to the constraint:

$\displaystyle g(x,y)=x+y-\frac{\pi}{2}=0$

$\displaystyle \cos(x)=\lambda=-\cos(y)$

$\displaystyle \cos(x)+\cos(y)=0$

Using a sum to product identity, and dividing through by $2$, we obtain:

$\displaystyle \cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=0$

On the given domain, this gives us:

$\displaystyle x\pm y=\pi\implies x=\pi\pm y$

Putting this into the constraint, we find:

$\displaystyle \pi\pm y+y=\frac{\pi}{2}\implies y=-\frac{\pi}{4}\implies x=\frac{3\pi}{4}$

Thus, the objective function at this critical point is:

$\displaystyle f\left(\frac{3\pi}{4},-\frac{\pi}{4}\right)=\sqrt{2}$

At the end-points of the given domain, we find:

$\displaystyle f\left(\frac{\pi}{2},0\right)=1$

$\displaystyle f\left(\pi,-\frac{\pi}{2}\right)=1$

And so we conclude that on the given domain, we have:

$\displaystyle f_{\min}=1$

$\displaystyle f_{\max}=\sqrt{2}$

2. We can also do a classical function analysis for$f(x)=\sin x - \cos x$ on the domain $[\frac\pi 2,\pi]$.
To find the extrema:
$$f'(x)=0 \quad\Rightarrow\quad \cos x + \sin x = 0 \quad\Rightarrow\quad \sin x=-\cos x \quad\Rightarrow\quad \tan x=-1\quad\Rightarrow\quad x=\frac{3\pi}{4}$$
So we have one extremum at $x=\frac{3\pi}{4}$ with value $\sqrt 2$.
And we have 2 boundary extrema at $x=\frac{\pi}{2}$ respectively $x=\pi$, both with value $1$.
Therefore:
$$1 \le \sin x - \cos x \le \sqrt 2 \quad\text{ when } x\in [\frac\pi 2,\pi]$$

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