# Showing that $\dfrac{\sin \alpha-\sin \beta}{\cos \beta-\cos \alpha}=\cot \theta$ for some $\theta$ between $\alpha$ and $\beta$ [closed]

I am stuck on the following problem that says:

Show that $$\frac{\sin \alpha-\sin \beta}{\cos \beta-\cos \alpha}=\cot \theta$$ where $0 \lt \alpha \lt \theta \lt \beta \lt \frac{\pi}{2}$.

## closed as off-topic by TomGrubb, Saad, Xander Henderson, Kavi Rama Murthy, ShaileshMay 29 '18 at 9:05

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• What have you tried? This also seems evidently false, unless some other assumption has been omitted, since cotangent is surely not constant on the open interval $(a, b)$. – Alex Nolte May 28 '18 at 19:57
• If $\alpha = \pi/10, \theta = \pi/4, \beta = \pi/3$ this seems to be false. – Gibbs May 28 '18 at 20:03

Apply Cauchy's mean value theorem:

If $f$ and $g$ are continuous on the closed interval $[\alpha,\beta]$, differentiable on its interior $(\alpha,\beta)$, there exists $\theta \in (\alpha,\beta)$ such that $$\frac{f(\alpha)-f(\beta)}{g(\alpha)-g(\beta)}=\frac{f'(\theta)}{g'(\theta)}.$$

• That's what I wanted. – learner May 28 '18 at 21:03
• Very nice answer (+1). – the_candyman May 28 '18 at 21:55

Notice that:

$$\sin(\alpha) - \sin(\beta) = 2\sin\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)$$

$$\cos(\beta) - \cos(\alpha) = 2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$

Their ratio is:

$$\cot(\theta) = \displaystyle\frac{\cos\left(\displaystyle\frac{\alpha+\beta}{2}\right)}{\sin\left(\displaystyle\frac{\alpha+\beta}{2}\right)},$$

where $\theta = \displaystyle\frac{\alpha+\beta}{2}.$

• Thanks a lot ..Got it.. – learner May 28 '18 at 20:02
• Very cute! (+1) – Bernard May 28 '18 at 20:53