Монгол Бодлогын Сан

Заавар: $(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$-г авч үз.

Бодолт: \begin{align*} 1^2+(\sqrt2)^2&=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\\ &=\cos^2\alpha+2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha+2\sin\alpha\sin\beta+\sin^2\beta\\ &=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)+2\cos\alpha\cos\beta+2\sin\alpha\sin\beta\\ &=2+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=2+2\cos(\alpha-\beta) \end{align*} тул $$3=2+2\cos(\alpha-\beta)\Rightarrow\cos(\alpha-\beta)=\dfrac12$$ байна.