# complex trigo

• February 7th 2011, 12:51 AM
prasum
complex trigo
the value of 169e^(pi+sin(inverse)(12/13)+cos(inverse)(5/13))

i know de moivres theorem cosa+isina

but how to proceed with inverses
• February 7th 2011, 01:13 AM
emakarov
Is there $i=\sqrt{-1}$ somewhere in this formula?
• February 7th 2011, 03:46 AM
HallsofIvy
Quote:

Originally Posted by prasum
the value of 169e^(pi+sin(inverse)(12/13)+cos(inverse)(5/13))

i know de moivres theorem cosa+isina

but how to proceed with inverses

With no "i" Demoivre's theorem does not apply. I see nothing to do but evaluate that using a calculator.

If that was supposed to be $169e^{i(\pi+ sin^{-1}(5/13)+ cos^{-1}(12/13))}$
then use the fact that $e^{a+ b}= e^{a}e^b$

Of course, $e^{i sin^{-1}(5/13)}= cos(sin^{-1}(5/13))+ i sin(sin^{-1}(5/13))$. You should recognize that $sin^{-1}(5/13)$ refers to an angle in a right triangle with "opposite side" 5 and "hypotenuse" 13. What is the length of the other leg? So what are the sine and cosine?
• February 7th 2011, 04:58 AM
Soroban
Hello, prasum!

Quote:

$\text}Find the value of: }\:169e^{\left(\pi+\sin^{\text{-}1}\!\frac{12}{13}+\cos^{\text{-}1}\!\frac{5}{13}\right)}$

$\text{Examine that exponent: }\:\pi + \underbrace{\sin^{\text{-}1}\!\!\left(\tfrac{12}{13}\right)}_{\alpha} + \underbrace{\cos^{\text{-}1}\!\!\left(\tfrac{5}{13}\right)}_{\beta}$

$\text{We have: }\:\begin{Bmatrix}\alpha \:=\:\sin^{-1}\frac{12}{13} & \Rightarrow & \sin\alpha \:=\:\frac{12}{13} \\ \\[-3mm]
\beta \:=\:\cos^{-1}\frac{5}{13} & \Rightarrow & \cos\beta \:=\:\frac{5}{12} \end{Bmatrix}$

Note that $\alpha$ and $\beta$ are the same angle!

Code:

             *           **           * *       13 *  * 12         *  *       *θ  *       *******         5 

Hence, the exponent is: . $\pi + \alpha + \alpha \:=\:\pi + 2\sin^{\text{-}1}\frac{12}{13}$

But what is the point of all this?

We could have cranked out the answer on our calculator.