There is a number of important relationships between the values of the functions $\sin$ and $\cos$. These are known as trigonometric identities. There are three of them which you should memorize, and about a dozen others which are less important.

The trigonometric functions are defined as $$\cos(\theta)=x_P~~,~~\sin(\theta)=y_P~~,~~\tan(\theta)=\frac{y_P}{x_P},$$ where $P=(x_P,y_P)$ is a point on the unit circle.

The three identities that you must remember are:

$$\sin^2(x)+\cos^2(x)=1.$$ This is true by Pythagoras theorem and the definition of sin and cos. The ratios of the squares of the sides of a triangle is equal to the square of the size of the hypotenuse.

$$\sin(a + b)=\sin(a)\cos(b) + \sin(b)\cos(a).$$ The mnemonic for this one is “sico sico”.

$$\cos(a + b)=\cos(a)\cos(b) - \sin(a)\sin(b).$$ The mnemonic for this one is “coco - sisi”—the negative sign is there because it is not good to be a sissy.

If you remember the above thee formulas, you can derive pretty much all the other trigonometric identities.

Starting from the sico-sico identity above, and setting $a=b=x$ we can derive following identity: $$\sin(2x) = 2\sin(x)\cos(x).$$

Starting from the coco-sisi identity, we derive: $$\cos(2x) \ =\ 2\cos^2(x) - 1 \ = 2\left(1 - \sin^2(x)\right) - 1 = 1 - 2\sin^2(x),$$ or if we rewrite to isolate the $\sin^2$ and $\cos^2$ we get: $$\cos^2(x) = \frac{1}{2}\left(1+\cos(2x)\right), \qquad \sin^2(x) = \frac{1}{2}\left(1-\cos(2x)\right).$$

Sin and cos are periodic functions with period $2\pi$. So if we add multiples of $2\pi$ to the input, we get the same value: $$\sin(x + 2\pi)=\sin(x +124\pi) = \sin(x), \qquad \cos(x+2\pi)=\cos(x).$$

Furthermore, sin and cos are self similar within each $2\pi$ cycle: $$\sin(\pi-x)=\sin(x), \qquad \cos(\pi-x)=-\cos(x).$$

Now it should come and no surprise if I tell you that actually sin and cos are just $\frac{\pi}{2}$-shifted versions of each other: $$\cos(x)=\sin\!\left(x\!+\!\frac{\pi}{2}\right)=\sin\!\left(\frac{\pi}{2}\!-\!x\right), \ \ \sin\!\left(x\right) = \cos\left(x\!-\!\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\!-\!x\right).$$

$$\sin\!\left(a\right)+\sin\!\left(b\right)=2\sin\!\left(\frac{1}{2}(a+b)\right)\cos\!\left(\frac{1}{2}(a-b)\right),$$ $$\sin\!\left(a\right)-\sin\!\left(b\right)=2\sin\!\left(\frac{1}{2}(a-b)\right)\cos\!\left(\frac{1}{2}(a+b)\right),$$ $$\cos\!\left(a\right)+\cos\!\left(b\right)=2\cos\!\left(\frac{1}{2}(a+b)\right)\cos\!\left(\frac{1}{2}(a-b)\right),$$ $$\cos\!\left(a\right)-\cos\!\left(b\right)=-2\sin\!\left(\frac{1}{2}(a+b)\right)\sin\!\left(\frac{1}{2}(a-b)\right).$$

$$\sin(a)\cos(b) = {1\over 2}(\sin{(a+b)}+\sin{(a-b)}),$$ $$\sin(a)\sin(b) = {1\over 2}(\cos{(a-b)}-\cos{(a+b)}),$$ $$\cos(a)\cos(b) = {1\over 2}(\cos{(a-b)}+\cos{(a+b)}).$$

The above formulas will come in handy in many situations when you have to find some unknown in an equation or when you are trying to simplify a trigonometric expression. I am not saying you should necessarily memorize them, but you should be aware that they exist.