Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if \[ ab\cos(\alpha - \beta) \leq \sqrt{(1-a^2)(1-b^2)}, \]then \[ a\cos\alpha + b\sin\beta \leq 1 + ab\sin(\beta - \alpha). \]
Problem
Source: Nordic Mathematical Contest 2017 Problem 2
Tags: trigonometry, inequalities
RagvaloD
04.04.2017 14:27
$1-a^2-b^2+a^2b^2 \geq a^2b^2\cos^2(\alpha - \beta)=a^2b^2(1-\sin^2(\alpha - \beta)) =a^2b^2-a^2b^2\sin^2(\alpha - \beta)$ $a^2b^2\sin^2(\alpha - \beta) \geq a^2+b^2-1$ $(1-ab \sin(\alpha- \beta))^2=1+a^2b^2\sin^2(\alpha - \beta)-2ab\sin(\alpha - \beta) \geq a^2+b^2-2ab \sin(\alpha - \beta)$ $ (a\cos\alpha + b\sin\beta)^2=a^2 \cos^2\alpha+b^2\sin^2\beta+2ab \cos\alpha \sin\beta=a^2+b^2-a^2\sin^2\alpha-b^2\cos^2\beta+2ab \cos\alpha \sin\beta=a^2+b^2-(a\sin\alpha+b\cos\beta)^2+2ab\sin \alpha\cos \beta+2ab \cos\alpha \sin\beta=a^2+b^2-(a\sin\alpha+b\cos\beta)^2+2ab\sin (\alpha+ \beta)=a^2+b^2-2ab \sin(\alpha - \beta) -(a\sin\alpha+b\cos\beta)^2+2ab \sin \alpha \cos \beta = a^2+b^2-2ab \sin(\alpha - \beta) -(a\sin\alpha-b\cos\beta)^2 \leq a^2+b^2-2ab \sin(\alpha - \beta) \leq (1-ab \sin(\alpha- \beta))^2$
CactusYang
19.06.2018 15:48
RagvaloD wrote: $1-a^2-b^2+a^2b^2 \geq a^2b^2\cos^2(\alpha - \beta)=a^2b^2(1-\sin^2(\alpha - \beta)) =a^2b^2-a^2b^2\sin^2(\alpha - \beta)$ $a^2b^2\sin^2(\alpha - \beta) \geq a^2+b^2-1$ $(1-ab \sin(\alpha- \beta))^2=1+a^2b^2\sin^2(\alpha - \beta)-2ab\sin(\alpha - \beta) \geq a^2+b^2-2ab \sin(\alpha - \beta)$ $ (a\cos\alpha + b\sin\beta)^2=a^2 \cos^2\alpha+b^2\sin^2\beta+2ab \cos\alpha \sin\beta=a^2+b^2-a^2\sin^2\alpha-b^2\cos^2\beta+2ab \cos\alpha \sin\beta=a^2+b^2-(a\sin\alpha+b\cos\beta)^2+2ab\sin \alpha\cos \beta+2ab \cos\alpha \sin\beta=a^2+b^2-(a\sin\alpha+b\cos\beta)^2+2ab\sin (\alpha+ \beta)=a^2+b^2-2ab \sin(\alpha - \beta) -(a\sin\alpha+b\cos\beta)^2+2ab \sin \alpha \cos \beta = a^2+b^2-2ab \sin(\alpha - \beta) -(a\sin\alpha-b\cos\beta)^2 \leq a^2+b^2-2ab \sin(\alpha - \beta) \leq (1-ab \sin(\alpha- \beta))^2$ The last paragraph should be: $4ab\sin\alpha\cos\beta$ before the last "$=$" mark.