Answer :
To solve the given expression [tex]\(\frac{6\left[\cos \left(70^{\circ}\right)+i \sin \left(70^{\circ}\right)\right]}{2\left[\cos \left(5^{\circ}\right)+i \sin \left(5^{\circ}\right)\right]}\)[/tex], follow these steps:
1. Recall Euler's Formula:
Using Euler's formula [tex]\( e^{i\theta} = \cos(\theta) + i\sin(\theta) \)[/tex], we can rewrite the expression.
[tex]\[ 6 \left[\cos(70^\circ) + i\sin(70^\circ)\right] = 6e^{i70^\circ} \][/tex]
[tex]\[ 2 \left[\cos(5^\circ) + i\sin(5^\circ)\right] = 2e^{i5^\circ} \][/tex]
2. Simplify the Fraction:
Now divide the two complex numbers:
[tex]\[ \frac{6e^{i70^\circ}}{2e^{i5^\circ}} = 3 \cdot \frac{e^{i70^\circ}}{e^{i5^\circ}} \][/tex]
Recall that [tex]\(\frac{e^{ia}}{e^{ib}} = e^{i(a-b)}\)[/tex]:
[tex]\[ 3 \cdot e^{i(70^\circ - 5^\circ)} = 3 \cdot e^{i65^\circ} \][/tex]
3. Convert Back to Trigonometric Form:
Rewriting [tex]\( e^{i65^\circ} \)[/tex] in trigonometric form:
[tex]\[ e^{i65^\circ} = \cos(65^\circ) + i\sin(65^\circ) \][/tex]
4. Combine the Results:
Combining the magnitude and the angle, the expression simplifies to:
[tex]\[ 3 \left[\cos(65^\circ) + i \sin(65^\circ)\right] \][/tex]
Therefore, the correct answer to [tex]\(\frac{6\left[\cos \left(70^{\circ}\right)+i \sin \left(70^{\circ}\right)\right]}{2\left[\cos \left(5^{\circ}\right)+i \sin \left(5^{\circ}\right)\right]}\)[/tex] is:
[tex]\[ 3 \left[\cos(65^\circ) + i \sin(65^\circ)\right] \][/tex]
Hence, the choice that matches this result is:
[tex]\[ \boxed{3\left[\cos \left(65^{\circ}\right)+i \sin \left(65^{\circ}\right)\right]} \][/tex]
1. Recall Euler's Formula:
Using Euler's formula [tex]\( e^{i\theta} = \cos(\theta) + i\sin(\theta) \)[/tex], we can rewrite the expression.
[tex]\[ 6 \left[\cos(70^\circ) + i\sin(70^\circ)\right] = 6e^{i70^\circ} \][/tex]
[tex]\[ 2 \left[\cos(5^\circ) + i\sin(5^\circ)\right] = 2e^{i5^\circ} \][/tex]
2. Simplify the Fraction:
Now divide the two complex numbers:
[tex]\[ \frac{6e^{i70^\circ}}{2e^{i5^\circ}} = 3 \cdot \frac{e^{i70^\circ}}{e^{i5^\circ}} \][/tex]
Recall that [tex]\(\frac{e^{ia}}{e^{ib}} = e^{i(a-b)}\)[/tex]:
[tex]\[ 3 \cdot e^{i(70^\circ - 5^\circ)} = 3 \cdot e^{i65^\circ} \][/tex]
3. Convert Back to Trigonometric Form:
Rewriting [tex]\( e^{i65^\circ} \)[/tex] in trigonometric form:
[tex]\[ e^{i65^\circ} = \cos(65^\circ) + i\sin(65^\circ) \][/tex]
4. Combine the Results:
Combining the magnitude and the angle, the expression simplifies to:
[tex]\[ 3 \left[\cos(65^\circ) + i \sin(65^\circ)\right] \][/tex]
Therefore, the correct answer to [tex]\(\frac{6\left[\cos \left(70^{\circ}\right)+i \sin \left(70^{\circ}\right)\right]}{2\left[\cos \left(5^{\circ}\right)+i \sin \left(5^{\circ}\right)\right]}\)[/tex] is:
[tex]\[ 3 \left[\cos(65^\circ) + i \sin(65^\circ)\right] \][/tex]
Hence, the choice that matches this result is:
[tex]\[ \boxed{3\left[\cos \left(65^{\circ}\right)+i \sin \left(65^{\circ}\right)\right]} \][/tex]