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Transforming the complex number $z=-\sqrt{3}+3i$ into polar form will bring me to the problem to solve this two equations to find the angle $\phi$: $\cos{\phi}=\frac{\Re z}{|z|}$ and $\sin{\phi}=\frac{\Im z}{|z|}$.

For $z$ the solutions are $\cos{\phi}=-0,5$ and $\sin{\phi}=-0,5*\sqrt{3}$. Using Wolfram Alpha or my calculator I can get $\phi=\frac{2\pi}{3}$ as solution. But using a calculator is forbidden in my examination.

Do you know any (cool) ways to get the angle without any other help?

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    There are **very few** instances where the answer will be "nice." One needs to learn them. And you probably did at some stage. Maybe the reason you did not recognize it is writing $0.5$ instead of $\frac{1}{2}$. If you had been looking for $\cos \phi=-\frac{1}{2}$, $\sin\phi=\frac{\sqrt{3}}{2}$ you might have remembered.2012-07-16

3 Answers 3

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memorize sin/cos for angles $0,{\pi \over 6},{\pi \over 4},{\pi \over 3},{\pi \over 2}$ and in your examination look at the unit circle to figure out what is going on

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    memorize $\sin \phi$ and calculate $\cos \phi = \sin (\phi +\pi/2)$2012-07-16
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You could start with known angles (s.a. multiples of 45 or 30 degrees) and work your way from there using the formulas for trigonometric half-angles, and sums of angles. If you don't remember them, use: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

For instance, in degrees, if you want cos(41), you can use the sequence: 45+120=165 165/2=82.5 82.5/2=41.25

and use the trigonometric identities to fall back from cos(45) and cos(120) to the approximation cos(41.25)

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The direct calculation is $\arg(-\sqrt 3+ 3i)=\arctan\frac{3}{-\sqrt{3}}=\arctan (-\sqrt 3)=\arctan \frac{\sqrt 3/2}{-1/2}$

As the other two answers remark, you must learn by heart the values of at least the sine and cosine at the main angle values between zero and $\,\pi/2\,$ and then, understanding the trigonometric circle, deduce the functions' values anywhere on that circle.

The solution you said you got is incorrectly deduced as you wrote $\,cos\phi=-0,5\,,\,\sin\phi=-0,5\cdot \sqrt 3\,$ which would give you both values of $\,\sin\,,\,\cos\,$ negative, thus putting you in the third quadrant of the trigonometric circle, $\,\{(x,y)\;:\;x,y<0\}\,$, which is wrong as the value indeed is $\,2\pi/3\,$ (sine is positive!), but who knows how did you get to it.

In the argument's calculation above please do note the minus sign is at $\,1/2\,$ in the denominator, since that's what corresponds to the $\,cos\,$ in the polar form, but the sine is positive thus putting us on the second quadrant $\,\{(x,y,)\;:\;x<0 .

So knowing that $\sin x = \sin(\pi - x)\,,\,\cos x=-\cos(\pi -x)\,,\,0\leq x\leq \pi/2$ and knowing the basic values for the basic angles, gives you now $-\frac{1}{2}=-\cos\frac{\pi}{3}\stackrel{\text{go to 2nd quad.}}=\cos\left(\pi-\frac{\pi}{3}\right)=\cos\frac{2\pi}{3}$ $\frac{\sqrt 3}{2}=\sin\frac{\pi}{3}\stackrel{\text{go to 2nd quad.}}=\sin\left(\pi-\frac{\pi}{3}\right)=\sin\frac{2\pi}{3}$