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DrunkenOldFool
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- Feb 6, 2012
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If $\alpha>0$, $\beta< \pi$ and $\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2$, then what is the value of $\sin(\alpha)+\cos(\beta)$?
Hi DrunkenOldFool,If $\alpha>0$, $\beta< \pi$ and $\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2$, then what is the value of $\sin(\alpha)+\cos(\beta)$?
Hi DrunkenOldFool,Thank You Sudharaka! Can you suggest any other simpler method?(f)
The solution $\alpha=(2n+1)\pi$ does not work, because in that case it would follow that $\sin\alpha=0$. The equation $\sin(\alpha)+\cos(\beta)=\frac{5}{4}$ then implies that $\cos(\beta)=\frac{5}{4}$, which is not possible.Hi DrunkenOldFool,
What I have shown you is that if you are given, \(\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2\) the value of \(\sin(\alpha)+\cos(\beta)\) will take different values depending on the value you choose for \(\alpha\). If you choose \(\alpha=(2n+1)\pi\mbox{ where }n\in\mathbb{Z}\) you have a corresponding \(\beta\) value which you can find from the equation given. For this \(\alpha\) and \(\beta\) values we have \(\sin(\alpha)+\cos(\beta)=\frac{5}{4}\). Similarly for \(\alpha=\frac{1}{3}(6n+1)\pi\mbox{ where }n\in\mathbb{Z}\) and the corresponding \(\beta\) value you have \(\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}\).
Hi Opalg,The solution $\alpha=(2n+1)\pi$ does not work, because in that case it would follow that $\sin\alpha=0$. The equation $\sin(\alpha)+\cos(\beta)=\frac{5}{4}$ then implies that $\cos(\beta)=\frac{5}{4}$, which is not possible.
It seems that the only solution with $\alpha$ and $\beta$ lying between 0 and $\pi$ is $\alpha=\beta=\pi/3$, in which case $\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}.$