About the sum of sines of two angles

Question

Suppose that $0\le \alpha\le \pi/2$ and $0\le \beta\le \pi/2$ such that $\alpha+\beta\ge \pi/2$. Can we prove that $\sin(\alpha)+\sin(\beta)\ge 1$?

Answer 1

Rewrite it as $$\sin(\alpha)+\sin(\beta)= 2 \sin\left( \frac{\alpha + \beta}{2} \right) \cos\left( \frac{\alpha - \beta}{2} \right)$$ Note that $\frac\pi2 \leq\alpha+\beta\leq\pi$ so that $\sin\left( \frac{\alpha + \beta}{2} \right)\geq\frac{1}{\sqrt2}$, and $-\frac\pi2\leq\alpha-\beta\leq\frac\pi2$ so that $\cos\left( \frac{\alpha - \beta}{2}\right)\geq \frac1{\sqrt2}$. Putting this together, we have $$\sin(\alpha)+\sin(\beta)= 2 \sin\left( \frac{\alpha + \beta}{2} \right) \cos\left( \frac{\alpha - \beta}{2} \right)\geq2\cdot\frac1{\sqrt2}\cdot\frac1{\sqrt2}=1$$

Answer 2

$$\sin\alpha+\sin\beta\ge \sin(\pi/2-\beta)+\sin\beta=\cos\beta+\sin\beta=\sqrt{\cos^2\beta+\sin^2\beta+2\cos\beta\sin\beta}\ge\sqrt 1=1$$

$0\le\pi/2-\beta\le\alpha\le\pi/2\implies\sin(\pi-\beta)\le\sin\alpha\\0\le\beta\le\pi/2\implies 0\le\cos\beta\land0\le\sin\beta$