Question

Without using trigonometric tables prove $\sin {40}^o-\cos {70}^o=\sqrt{3}\cos {80}^o$.

Answer 1

$$\begin{array}{c}\sin {40}^0-\cos \left({40}^0+{30}^0\right)\\ \sin {40}^0-\left[\cos {40}^0\cdot \cos {30}^0-\sin {40}^0\cdot \sin {30}^0\right]\\ \left[\sin {40}^0-\cos {40}^0\cdot \frac{\sqrt{3}}{2}+\sin {40}^0\cdot \frac{1}{2}\right]\\ \left[\frac{3}{2}\sin {40}^0-\cos {40}^0\cdot \frac{\sqrt{3}}{2}\right]\\ \sqrt{3}\left[\frac{\sqrt{3}}{2}\sin {40}^0-\cos {40}^0\cdot \frac{1}{2}\right]\\ \sqrt{3}\left[\sin {60}^0\cdot \sin {40}^0-\cos {40}^0\cdot \cos {40}^0\right]\\ \sqrt{3}\left[-\cos ({40}^0+{60}^0)\right]\\ \sqrt{3}\left[-\cos {100}^0\right]\\ \sqrt{3}\left[\cos ({180}^0-{100}^0)\right]\\ \sqrt{3}\cos {80}^0\end{array}$$