Proof verification: $\forall x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) \geq 1$.

Question

Would someone be willing to verify the following proof by contradiction?

Theorem: $\forall x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) \geq 1$.

Suppose, for the sake of contradiction, that this statement is not true. Then, $\exists x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) < 1$.

$sin(x) - cos(x) < 1 \Longrightarrow sin^2(x) - 2sin(x)cos(x) + cos^2(x) < 1$

$\Longrightarrow sin^2(x) + cos^2(x) < 1 + 2sin(x)cos(x) \Longrightarrow 0 < sin(x)cos(x)$

On the interval $(\frac{\pi}{2}, \pi), sin(x) > 0$, and $cos(x) < 0$. Therefore, $sin(x)cos(x) < 0$, which is a contradiction.

Answer 1

Use that

$$\cos x=\sin \left( \frac{\pi}{2} -x\right)$$

and the sum-to-product formulas.

Notably from

$$\sin \theta - \sin \varphi = 2 \sin\left( \frac{\theta - \varphi}{2} \right) \cos\left( \frac{\theta + \varphi}{2} \right)$$

we obtain

$$\sin x - cos x=\sin x -\sin \left( \frac{\pi}{2} -x\right)=2\sin\left(x-\frac{\pi}{4}\right) \cdot\cos \frac{\pi}{4}=\sqrt2\cdot \sin\left(x-\frac{\pi}{4}\right) \geq 1$$

whic is true since

$$\frac{\pi}{2}\le x\le\pi\iff\frac{\pi}{4}\le x-\frac{\pi}{4}\le\frac{3\pi}{4}\implies \sin\left(x-\frac{\pi}{4}\right)\ge\frac{\sqrt2}{2} $$

Answer 2

In order to square both sides of an equality we need to make sure that both sides are positive.

Thus if you mention that $ \forall x \in [\frac{\pi}{2}, \pi]$ , $$ 0< sin(x) - cos(x) < 1 \Longrightarrow sin^2(x) - 2sin(x)cos(x) + cos^2(x) < 1$$ Then your proof is flawless.

Answer 3

I prefer the following approach: We know $$\sin(a+b)=\sin a\cos b+\cos a\sin b.$$ Look for $c$ such that $c\sin x-c\cos x=\sin(x+a)$ for some $a$. You need $c=\pm\sqrt2/2$. To make life easier, take $c>0$. This gives us $a=-\pi/4$, and $x+a\in[\pi/4,3\pi/4]$. Proceed from here.