# How do you express sin(pi/ 6 ) * cos( ( 3 pi) / 4 )  without using products of trigonometric functions?

Using Sum or Difference
$\frac{1}{2} \cdot \sin \left(\frac{11 \pi}{12}\right) - \frac{1}{2} \cdot \sin \left(\frac{7 \pi}{12}\right)$

#### Explanation:

From "Sum and Difference Formulas"

$\sin \left(x + y\right) = \sin x \cos y + \cos x \sin y$
$\sin \left(x - y\right) = \sin x \cos y - \cos x \sin y$

$\sin \left(x + y\right) + \sin \left(x - y\right) = 2 \cdot \sin x \cos y$

so that, after dividing both sides by 2

$\sin x \cos y = \frac{1}{2} \cdot \sin \left(x + y\right) + \frac{1}{2} \cdot \sin \left(x - y\right)$

at this point , Let $x = \frac{\pi}{6}$ and $y = \frac{3 \pi}{4}$

$\sin x \cos y = \frac{1}{2} \cdot \sin \left(x + y\right) + \frac{1}{2} \cdot \sin \left(x - y\right)$

$\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{3 \pi}{4}\right) = \frac{1}{2} \cdot \sin \left(\frac{\pi}{6} + \frac{3 \pi}{4}\right) + \frac{1}{2} \cdot \sin \left(\frac{\pi}{6} - \frac{3 \pi}{4}\right)$

$\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{3 \pi}{4}\right) = \frac{1}{2} \cdot \sin \left(\frac{11 \pi}{12}\right) + \frac{1}{2} \cdot \sin \left(\frac{- 7 \pi}{12}\right)$

but $\sin \left(\frac{- 7 \pi}{12}\right) = - \sin \left(\frac{7 \pi}{12}\right)$

therefore

$\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{3 \pi}{4}\right) = \frac{1}{2} \cdot \sin \left(\frac{11 \pi}{12}\right) - \frac{1}{2} \cdot \sin \left(\frac{7 \pi}{12}\right)$

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