Sin -3-A - cos -6-A -

The correct option is A √(3)cos x sin( π #160; 3 +A ) +cos( π #160; 6 +A ) =sin( π #160; 3 +A ) +sin( π #160; 2 π #160; 6 ...

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Sin(A+B)Sin(A-B)

sin π3+A + co...

Question

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

\sqrt{3} cos x

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\sqrt{2} cos x

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cos x

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None of theabove

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Solution

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{⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \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}

(cos \frac{π}{4} - sin \frac{π}{6}) (\frac{1}{cos x} + tan x) = sin \frac{π}{4} cos x

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

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)

A = [\begin{matrix} cos \frac{π}{3} & sin \frac{π}{3} sin \frac{π}{3} & cos \frac{π}{3} \end{matrix}]

A^{3} = ?

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