Prove that - cos -6-A- cos -3-B- sin -6-A- sin -3-B-cosA-B

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Cos(A+B)Cos(A-B)

Prove that : ...

Question

Prove that : $c o s (\frac{π}{6} - A) \cdot c o s (\frac{π}{3} + B) - s i n (\frac{π}{6} - A) \cdot s i n (\frac{π}{3} + B) = cos (A - B)$

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sin (\frac{π}{6} + A) \cdot cos (\frac{π}{3} - B) + sin (\frac{π}{3} - B) \cdot cos (\frac{π}{6} + A) = cos A - B

cos (\frac{π}{6} - A) cos (\frac{π}{3} + B) - sin (\frac{π}{6} - A) sin (\frac{π}{3} + B) =

sin [\frac{π}{6} + A] . cos [\frac{π}{3} - B] + sin [\frac{π}{3} - B] . cos [\frac{π}{6} + A] = cos (A - B)

{⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{sin \frac{π}{6} + i (1 - cos \frac{π}{6})}{sin \frac{π}{6} - i (1 - cos \frac{π}{6})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{3}

i = \sqrt{- 1}

sin (\frac{π}{3} + A) + cos (\frac{π}{6} + A) =

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