Question

$ \frac{sin\,60^{\circ}+i\,cos\,60^{\circ}}{cos\,15^{\circ}-i\,sin\,15^{\circ}}= $

• A. $ \frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}} $
• B. $ \frac{1}{\sqrt{2}}-i \frac{1}{\sqrt{2}} $
• C. $ -\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}} $
• D. $ \frac{1}{2}+i \frac{\sqrt{3}}{2} $

Answer 1

Answer: (A)

We have,
$\frac{sin\,60^{\circ} + i\, cos\, 60^{\circ}}{cos\, 15^{\circ} - i \,sin \,15^{\circ}} \times \frac{cos\,15^{\circ} + i \,sin\,15^{\circ}}{cos\, 15^{\circ} + i\, sin\,15^{\circ}}$
$= \frac{(sin\,60^{\circ} cos \,15^{\circ} - cos 60^{\circ} sin 15^{\circ}) + i(cos 60^{\circ} cos 15^{\circ} + sin \,60^{\circ} sin\,15^{\circ})}{cos^{2} 15^{\circ} + sin^{2} 15^{\circ}}$
$ = sin (60^{\circ} - 15^{\circ}) + i \, cos (60^{\circ} - 15^{\circ})$
$= sin\,45^{\circ} + i \,cos\,45^{\circ}= \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$