# Remember trig identities like $\cos(\pi/3) = 1/2$

I have started doing complex analysis and I keep having to switch between rectangular co-ordinates and polar form and I keep running into stuff like - $\cos(\pi/3) = 1/2$. I keep having to look these up. Am I expected to memorize these or what, its getting to be a serious pain having to keep looking them up. Do people generally just memorize these identities for complex analysis?

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You may wish to fix this: $$\cos(\pi/3)=\sqrt{3}/2$$ It's actually 1/2: wolframalpha.com/input/?i=cos%28pi%2F3%29 . – 000 Feb 4 '12 at 3:28

(Image from Wikipedia Commons)

Check this picture out! Using the Pythagorean Theorem, you can find the length of the dotted line yourself, so finding $\sin(60)$ and $\sin(30)$ will never be more than a triangle away.

By drawing a 45-45-90 triangle, you can easily get $\sin(45)$ yourself too.

Don't worry about the memorization part: after using these enough, trust me, you won't have to look them up!

(Oh, and if the need arises, you can also get $\sin(15)$ and $\sin(75)$ yourself with the addition formulas. But I've never really used these or even bothered to try to memorize them.)

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Here's my song about the 30-60-90 triangle: math.stackexchange.com/a/99432/409 – Blue Feb 4 '12 at 0:15

People generally know $\cos(\pi/3)=1/2$ long before they get to complex analysis (but maybe you really mean complex numbers). Consider an equilateral triangle with sides of length 2. Cut it in half by dropping a perpendicular from one vertex to the opposite side. Each half is a 30-60-90 triangle with short side 1 and hypotenuse 2, and by Pythagoras the long side is $\sqrt3$. Now if you remember cosine is adjacent over hypotenuse, you can recover all the trig functions at $\pi/6$ and $\pi/3$.

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Here is a dumb but effective way to learn these things. Here we see the sines and cosines of the "key angles", 0, 30, 45, 60, 90.

$$\sin(0) = \sqrt{0/4} = \cos(\pi/2)$$

$$\sin(\pi/6) = \sqrt{1/4} = \cos(\pi/3)$$

$$\sin(\pi/4) = \sqrt{2/4}= \cos(\pi/4)$$

$$\sin(\pi/3) = \sqrt{3/4} = \cos(\pi/6)$$

$$\sin(\pi/2) = \sqrt{4/4} = \cos(0)$$

Also, remember that co'' in a trig function means complement. We have $${\rm trig}(\theta) = {\rm cotrig}(\pi/2 - \theta).$$ Note that $${\rm cocotrig}(\theta) = {\rm trig(\theta)}.$$

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All you really need to know are the values of $\sin(x)$, where $x$ is $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$, and $\frac{\pi}{2}$. All of the other values come directly from these:

\begin{align} \sin(0) &= \frac{\sqrt{0}}{2} = 0 \\ \\ \sin(\pi/6) &= \frac{\sqrt{1}}{2} = \frac{1}{2} \\ \\ \sin(\pi/4) &= \frac{\sqrt{2}}{2} \\ \\ \sin(\pi/3) &= \frac{\sqrt{3}}{2} \\ \\ \sin(\pi/2) &= \frac{\sqrt{4}}{2} = 1 \end{align}

These value are easy to memorize, or you can just find them using Gerry's or Lopsy's answers.

Now, say you want to find $\cos(\pi/3)$. Since $\sin(\pi/3) = \frac{\sqrt{3}}{2}$, and $\sin^2(x) + \cos^2(x) = 1$, you get that

$$\cos^2(\pi/3) = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4}.$$

Hence, $\cos(\pi/3) = \pm \frac{1}{2}$. Since $\pi/3$ is in the first quadrant, it must be the positive value.

This might be the long way around (and perhaps looking at triangles or the unit circle is easier), but this saves you from looking up any values.

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