$$ \frac{3 \operatorname{Cis} 60^{\circ} .2 \frac{\pi^{-18^{\circ}}}{\frac{30^{\circ}}{}}}{6 \operatorname{Cis} 50^{\circ}} $$ to rectangular form

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To convert the expression $$ \frac{3 \operatorname{Cis} 60^{\circ} \cdot 2 \frac{\pi^{-18^{\circ}}}{\frac{30^{\circ}}{}}}{6 \operatorname{Cis} 50^{\circ}} $$ to rectangular form, we will first simplify the expression step by step.

1. **Understanding the notation**: 
   - The notation $$ \operatorname{Cis} \theta $$ is shorthand for $$ \cos \theta + i \sin \theta $$.
   - Therefore, $$ \operatorname{Cis} 60^{\circ} = \cos 60^{\circ} + i \sin 60^{\circ} = \frac{1}{2} + i \frac{\sqrt{3}}{2} $$.
   - Similarly, $$ \operatorname{Cis} 50^{\circ} = \cos 50^{\circ} + i \sin 50^{\circ} $$.

2. **Simplifying the expression**:
   - The expression can be rewritten as:
   $$
   \frac{3 \cdot \operatorname{Cis} 60^{\circ} \cdot 2 \cdot \pi^{-18^{\circ}}}{6 \cdot \operatorname{Cis} 50^{\circ}}
   $$
   - This simplifies to:
   $$
   \frac{6 \cdot \operatorname{Cis} 60^{\circ} \cdot \pi^{-18^{\circ}}}{6 \cdot \operatorname{Cis} 50^{\circ}} = \frac{\operatorname{Cis} 60^{\circ} \cdot \pi^{-18^{\circ}}}{\operatorname{Cis} 50^{\circ}}
   $$

3. **Dividing the complex numbers**:
   - We can use the property of division of complex numbers in polar form:
   $$
   \frac{\operatorname{Cis} \theta_1}{\operatorname{Cis} \theta_2} = \operatorname{Cis}(\theta_1 - \theta_2)
   $$
   - Thus, we have:
   $$
   \frac{\operatorname{Cis} 60^{\circ}}{\operatorname{Cis} 50^{\circ}} = \operatorname{Cis}(60^{\circ} - 50^{\circ}) = \operatorname{Cis}(10^{\circ})
   $$

4. **Including the factor $$ \pi^{-18^{\circ}} $$**:
   - The expression now becomes:
   $$
   \operatorname{Cis}(10^{\circ}) \cdot \pi^{-18^{\circ}}
   $$

5. **Calculating $$ \pi^{-18^{\circ}} $$**:
   - The term $$ \pi^{-18^{\circ}} $$ is a bit unusual. If we interpret it as $$ e^{-i \cdot 18^{\circ}} $$, we can express it in rectangular form:
   $$
   \pi^{-18^{\circ}} = e^{-i \cdot 18^{\circ}} = \cos(-18^{\circ}) + i \sin(-18^{\circ}) = \cos(18^{\circ}) - i \sin(18^{\circ})
   $$

6. **Combining everything**:
   - Now we can express the final result:
   $$
   \operatorname{Cis}(10^{\circ}) \cdot (\cos(18^{\circ}) - i \sin(18^{\circ}))
   $$
   - This gives us:
   $$
   (\cos(10^{\circ}) + i \sin(10^{\circ})) \cdot (\cos(18^{\circ}) - i \sin(18^{\circ}))
   $$

7. **Using the distributive property**:
   - Expanding this product:
   $$
   = \cos(10^{\circ}) \cos(18^{\circ}) + \sin(10^{\circ}) \sin(18^{\circ}) + i (\sin(10^{\circ}) \cos(18^{\circ}) - \cos(10^{\circ}) \sin(18^{\circ}))
   $$

8. **Final rectangular form**:
   - The real part is:
   $$
   \cos(10^{\circ}) \cos(18^{\circ}) + \sin(10^{\circ}) \sin(18^{\circ}) = \cos(10^{\circ} - 18^{\circ}) = \cos(-8^{\circ}) = \cos(8^{\circ})
   $$
   - The imaginary part is:
   $$
   \sin(10^{\circ}) \cos(18^{\circ}) - \cos(10^{\circ}) \sin(18^{\circ}) = \sin(10^{\circ} - 18^{\circ}) = \sin(-8^{\circ}) = -\sin(8^{\circ})
   $$

Thus, the final rectangular form of the expression is:
$$
\cos(8^{\circ}) - i \sin(8^{\circ})
$$
The rectangular form is $$ \cos(8^{\circ}) - i \sin(8^{\circ}) $$.

to rectangular form

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