(sin 60°+i cos 60°/cos 15°-i sin 15°)=

Question

(sin 60°+i cos 60°/cos 15°-i sin 15°)=  (A)  (1/√2)+i (1/√2)  (B)  (1/√2)-i (1/√2)  (C)  -(1/√2)+i (1/√2)  (D)  (1/2)+i (√3/2)

Tardigrade Admin · Accepted Answer

We have, (sin 60° + i cos 60°/cos 15° - i sin 15°) × (cos 15° + i sin 15°/cos 15° + i sin 15°) = ((sin 60° cos 15° - cos 60° sin 15°) + i(cos 60° cos 15° + sin 60° sin 15°)/cos2 15° + sin2 15°) = sin (60° - 15°) + i cos (60° - 15°) = sin 45° + i cos 45°= (1/√2) + i (1/√2)

Q. $\frac{s in 6 0 ^{\circ} + i cos 6 0 ^{\circ}}{cos 1 5 ^{\circ} - i s in 1 5 ^{\circ}} =$

$\frac{1}{2} + i \frac{1}{2}$

$\frac{1}{2} - i \frac{1}{2}$

$- \frac{1}{2} + i \frac{1}{2}$

$\frac{1}{2} + i \frac{3}{2}$

Q. cos15∘−isin15∘sin60∘+icos60∘​=

2​1​+i2​1​

2​1​−i2​1​

−2​1​+i2​1​

21​+i23​​

We have, cos15∘−isin15∘sin60∘+icos60∘​×cos15∘+isin15∘cos15∘+isin15∘​ =cos215∘+sin215∘(sin60∘cos15∘−cos60∘sin15∘)+i(cos60∘cos15∘+sin60∘sin15∘)​ =sin(60∘−15∘)+icos(60∘−15∘) =sin45∘+icos45∘=2​1​+i2​1​

Q. $\frac{s in 6 0 ^{\circ} + i cos 6 0 ^{\circ}}{cos 1 5 ^{\circ} - i s in 1 5 ^{\circ}} =$

$\frac{1}{2} + i \frac{1}{2}$

$\frac{1}{2} - i \frac{1}{2}$

$- \frac{1}{2} + i \frac{1}{2}$

$\frac{1}{2} + i \frac{3}{2}$