Trigonometry Challenge 2022/09/26

Solution: Since $\alpha$, $\beta$, and $\gamma$ are the interior angles of $\triangle{ABC}$, we have $$0\lt\alpha,\ \beta,\ \gamma\lt\pi$$ Therefore $$0\lt\sin\alpha,\ \sin\beta,\ \sin\gamma\le 1$$ and $$\sin\alpha\cdot\sin\beta\cdot\sin\gamma\lt1$$

If $$\cos\alpha\cdot\cos\beta+\sin\alpha\cdot\sin\beta\cdot\sin\gamma=1 \tag{1}$$ then $$\cos\alpha\cdot\cos\beta\gt0$$ Therefore $$0\lt\alpha,\ \beta\le\dfrac{\pi}{2}$$ as $\cos\alpha$ and $\cos\beta$ cannot be both negative at the same time, which would result in $\dfrac{\pi}{2}\lt\alpha,\ \beta\lt\pi$ and $\gamma\lt 0$.

Since all terms in equation (1) are positive, and $\sin\gamma\le 1$, we have $$\cos\alpha\cdot\cos\beta+\sin\alpha\cdot\sin\beta\cdot\sin\gamma\le \cos\alpha\cdot\cos\beta+\sin\alpha\cdot\sin\beta\cdot=\cos(\alpha-\beta)$$

Since $\cos(\alpha-\beta)\le1$, we have $$\cos\alpha\cdot\cos\beta+\sin\alpha\cdot\sin\beta\cdot\sin\gamma\le1$$

The equal sign applies only if $\cos(\alpha-\beta)=1$ and $\sin\gamma=1$. This implies that the solution for equation (1) is $$\boxed{\alpha=\beta=\dfrac{\pi}{4},\ \gamma=\dfrac{\pi}{2}}$$ i.e. $\triangle{ABC}$ is an isosceles right triangle.