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?
[Math] Remember trig identities like $\cos(\pi/3) = 1/2$
complex-analysistrigonometry
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Everything starts with $$\sin(a+b)=\sin a\cos b+\cos a\sin b$$ This is an identity, it holds for all $a$ and $b$. In particular, you're allowed to replace $b$ with $a$, so long as you do it consistently throughout, and you get $$\sin2a=2\sin a\cos a$$ Stop me if you didn't follow this. Now we can replace $a$ everywhere with $2x$ and get $$\sin 4x=2\sin2x\cos2x$$ Now there's a $\sin2x$ in that formula; we can use double-angle on it to get $$\sin4x=2(2\sin x\cos x)\cos2x$$ Now multiplication is associative, which means as long as all we're doing is multiplication, we don't need parentheses. On the right side, we're multiplying 5 things: $$\sin4x=2\times2\times\sin x\times\cos x\times\cos2x$$ Finally, $2\times2=4$, so $$\sin4x=4\sin x\cos x\cos2x$$
I created the images that "inspired" the illustration of the Angle-Sum and Difference identities on Wikipedia, so I'm all in favor of learning those relations. However, this is how I remember your particular class of identities.
First, in the Unit Circle, I imagine the "sine-cosine-$1$" triangle for a generic first-quadrant angle $\theta$, taking $\theta$ small enough that the triangle is considerably wider than it is tall, creating an obvious "long leg" and "short leg". Because trig values are positive in the first quadrant, I can say that the "long leg" of the triangle has length $\cos\theta$, and that the "short leg" has length $\sin\theta$.
Then, I imagine what happens when I rotate that triangle through multiples of $90^\circ$, getting a nice windmill:
The "(co-)reference triangle" for each of the compound angles is a rotation of the original triangle, and we can read off sine and cosine values by paying attention to the positions of the short legs and long legs (and assigning signs, as appropriate). For instance, the point labeled "$\theta+90^\circ$" is at distance "$\cos\theta$" above the $x$-axis, and at distance "$\sin\theta$" to the left of the $y$-axis; therefore, $$\sin\left(\theta+\phantom{1}90^\circ\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta+\phantom{1}90^\circ\right) = -\color{blue}{\sin\theta}$$
Likewise, $$\sin\left(\theta+180^\circ\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\theta+180^\circ\right) = -\color{red}{\cos\theta}$$ $$\sin\left(\theta-\phantom{1}90^\circ\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta-\phantom{1}90^\circ\right) = \phantom{-}\color{blue}{\sin\theta}$$
Similarly, there's a windmill for compound angles involving $-\theta$:
And we have $$\sin\left(\phantom{180^\circ}-\theta\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\phantom{180^\circ}-\theta\right) = \phantom{-}\color{red}{\cos\theta}$$ $$\sin\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta}$$ $$\sin\left(180^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta} \qquad\qquad \cos\left(180^\circ-\theta\right) = -\color{red}{\cos\theta}$$ $$\sin\left(-90^\circ-\theta\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(-90^\circ-\theta\right) = -\color{blue}{\sin\theta}$$
It's worthwhile to point out that the identity $$\cos\theta = \sin\left(90^\circ - \theta\right)$$ is, for some (such as myself), definitional:
The co-sine of the angle is the sine of the co-angle.
where "co-angle" means "complementary angle", just as "co-sine" literally means "complementary sine".
Also, the identities
$$\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta$$
have significance in establishing that cosine is an "even function" (it acts on negative arguments ---killing the sign--- the way an even exponent would) and sine is an "odd function" (it acts on negative arguments ---preserving the sign--- the way an odd exponent would). These are handy properties, which say interesting things about how the graphs are drawn. So, you might want to reserve a special area of your brain-space for them.
Best Answer
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.)