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Mathematics Question on Inverse Trigonometric Functions

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

A

12+i12\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}

B

12−i12\frac{1}{\sqrt{2}}-i \frac{1}{\sqrt{2}}

C

−12+i12-\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}

D

12+i32\frac{1}{2}+i \frac{\sqrt{3}}{2}

Answer

12+i12\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}

Explanation

Solution

We have, sin 60∘+i cos 60∘cos 15∘−i sin 15∘×cos 15∘+i sin 15∘cos 15∘+i sin 15∘\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}} =(sin 60∘cos 15∘−cos60∘sin15∘)+i(cos60∘cos15∘+sin 60∘sin 15∘)cos215∘+sin215∘= \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∘−15∘)+i cos(60∘−15∘) = sin (60^{\circ} - 15^{\circ}) + i \, cos (60^{\circ} - 15^{\circ}) =sin 45∘+i cos 45∘=12+i12= sin\,45^{\circ} + i \,cos\,45^{\circ}= \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}