How do you evaluate $cos (\frac{π}{6}) cos (\frac{π}{3}) - sin (\frac{π}{6}) sin (\frac{π}{3})$ $cos (pi/6) cos (pi/3) - sin (pi/6) sin (pi/3)$ ?

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How do you evaluate $cos (\frac{π}{6}) cos (\frac{π}{3}) - sin (\frac{π}{6}) sin (\frac{π}{3})$ $cos (pi/6) cos (pi/3) - sin (pi/6) sin (pi/3)$ ?

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Answer 1 · 2016-03-25T12:00:51

Zero

Explanation:

Apply the trig identity:
cos (a + b) = cos a.cos b - sin a.sin b
$cos (\frac{π}{6}) cos (\frac{π}{3}) - sin (\frac{π}{3}) sin (\frac{π}{6}) = cos (\frac{π}{3} + \frac{π}{6}) = = cos (\frac{π}{2}) = 0$ $cos (pi/6)cos (pi/3) - sin (pi/3)sin (pi/6) = cos (pi/3 + pi/6) = = cos (pi/2) = 0$

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How do you evaluate $cos (\frac{π}{6}) cos (\frac{π}{3}) - sin (\frac{π}{6}) sin (\frac{π}{3})$ $cos (pi/6) cos (pi/3) - sin (pi/6) sin (pi/3)$ ?

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How do you evaluate $cos (\frac{π}{6}) cos (\frac{π}{3}) - sin (\frac{π}{6}) sin (\frac{π}{3})$ $cos (pi/6) cos (pi/3) - sin (pi/6) sin (pi/3)$ ?