Prove the following- sin - 6 -A -cos - 3 -B -sin - 3 -B -cos - 6 -A -cos A-B

As we know that, sinx.cosy+cosx.siny=sin(x+y) So using this identity, #160;we get LHS is equal to,= gt; sin(π6+A+π3 B) = gt; sin[π2+ (A B)] = ...

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

Prove the fol...

Question

Prove the following:
$sin (\frac{π}{6} + A) \cdot cos (\frac{π}{3} - B) + sin (\frac{π}{3} - B) \cdot cos (\frac{π}{6} + A) = cos A - B$

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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)

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

A + B + C = π

s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = 4 s i n \frac{π - 4}{4} . s i n \frac{π - B}{4} . s i n \frac{π - C}{4}

{⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \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}{6} cos 0 + sin \frac{π}{4} cos \frac{π}{4} + sin \frac{π}{3} cos \frac{π}{6} = \frac{7}{4}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

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