# Use identity to simplify the expression? 2sin(π/9-π/2)cos(π/2-π/9)

##### 1 Answer
Apr 2, 2018

$2 \sin \left(\frac{\pi}{9} - \frac{\pi}{2}\right) \cos \left(\frac{\pi}{2} - \frac{\pi}{9}\right) = - \sin \left(\frac{2 \pi}{9}\right)$

#### Explanation:

We use $\sin \left(- A\right) = - \sin A$, $\sin \left(\frac{\pi}{2} - A\right) = \cos A$, $\cos \left(\frac{\pi}{2} - A\right) = \sin A$ and $2 \sin A \cos A = \sin 2 A$

$\sin \left(\frac{\pi}{9} - \frac{\pi}{2}\right) = \sin \left(- \left(\frac{\pi}{2} - \frac{\pi}{9}\right)\right)$

= $- \sin \left(\frac{\pi}{2} - \frac{\pi}{9}\right) = - \cos \left(\frac{\pi}{9}\right)$

Similarly $\cos \left(\frac{\pi}{2} - \frac{\pi}{9}\right) = \sin \left(\frac{\pi}{9}\right)$

Hence $2 \sin \left(\frac{\pi}{9} - \frac{\pi}{2}\right) \cos \left(\frac{\pi}{2} - \frac{\pi}{9}\right)$

= $- 2 \cos \left(\frac{\pi}{9}\right) \cdot \sin \left(\frac{\pi}{9}\right)$

= $- 2 \sin \left(\frac{\pi}{9}\right) \cos \left(\frac{\pi}{9}\right)$

= $- \sin \left(\frac{2 \pi}{9}\right)$