Prove that e^ia + 2e^ib = re^ic?

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The discussion revolves around proving the equation e^ia + 2e^ib = re^ic, with given equations for r^2 and tan c. The user attempted to expand e^ia + 2e^ib to match r^2 but is confused about how to incorporate tan c into the proof. There is a question raised about the use of arctangent in this context. The conversation highlights the challenges in manipulating complex exponential forms and trigonometric identities. Clarification on these mathematical concepts is sought to complete the proof.
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Homework Statement


Prove that e^ia + 2e^ib = re^ic?



Homework Equations


given : r^2 = 5 + 4cos( a - b )
tan c = (sin a + 2 sin b) / (cos a + 2 cos b)
i is imaginary number


The Attempt at a Solution


I tried to expand e^ia + 2e^ib and make it equal to r^2
but I don't know what to do with the tan c
 
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tommykhoa said:
but I don't know what to do with the tan c
What's wrong with arctangent?
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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