## Cotangent Identity

$$\begin{array}{cc} & |\overline{AC}|\,|\overline{FD}| = |AEDF| = |\overline{AB}|\,|\overline{ED}| \\[6pt] \to \qquad & \large{b^2\,\left(\,\cot A + \cot B\,\right) = c^2\,\left(\,\cot A + \cot C\,\right)} \\ \end{array}$$

Posted 23 September, 2015 by in Proof Without Words, Trigonography

## Half-Angles in a Triangle

My answer to this question at Math.StackExchange.com. $$\Large{c\;\sin\frac{A-B}{2} \;\equiv\; (a-b)\;\cos\frac{C}{2}}$$

Posted 12 August, 2015 by in Proof Without Words, Trigonography

## Identity from Math.StackExchange.com

Taken from my answer to “Proving a weird trig identity” at Math.StackExchange.com . $$\Large{\frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta + \tan\theta}{1}}$$ In the diagram, $$\overline{AB}$$ is tangent to the unit circle at P. The “trig lengths” (except for $$|\overline{AQ}|$$) should be clear. We note that $$\angle BPR \cong \angle RPP^\prime$$, since these inscribed angles subtend congruent arcs $$\stackrel{\frown}{PR}$$ […]

Posted 14 May, 2015 by in Classroom, Trigonography

## Product Identities

\begin{align} 2 \cos A \cos B &= \cos(A-B) \;+\; \cos(A+B) \\ 2 \sin A \;\sin B &= \cos(A-B) \;-\; \cos(A+B) \end{align}

Posted 25 February, 2015 by in Classroom, Proof Without Words, Trigonography, Uncategorized

## Proof without words: Pi from Square(root)s

$$\huge \sqrt{2} + \sqrt{3} \;\approx\; \pi$$ \large \begin{align} \sqrt{2} + \sqrt{3} &= |\overline{AB}| + |\overline{CD}| \\ &= |\widehat{AB^\prime}| + |\widehat{CD^\prime}| \\ &\approx |\stackrel{\frown}{AB^\prime}| + |\stackrel{\frown}{CD^\prime}| \\ &\approx |\stackrel{\frown}{AC}| \\[6pt] &= \pi \end{align}

Posted 7 March, 2014 by in Uncategorized