You already know how to find the values of trigonometric functions for standard angles like 30°, 45°, and 60°. But what about an angle like 75°? We can't use a standard triangle for that. The trick is to express 75° as a sum of two familiar angles, like 45° + 30°. This is where the sum and difference identities come in.

Let's start with the core formulas for sine and cosine. All other compound angle formulas can be derived from these four.

Using these, we can find $\cos(75°)$. We can write $75°$ as $45° + 30°$.

$\cos(75°) = \cos(45° + 30°)$ $= \cos 45° \cos 30° - \sin 45° \sin 30°$

Substitute the known values: $= (\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) - (\frac{1}{\sqrt{2}})(\frac{1}{2})$ $= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$ $= \frac{\sqrt{3}-1}{2\sqrt{2}}$

What if the two angles are the same? Let's take the sum formula for sine, $\sin(A+B)$, and set $B = A$. This gives us $\sin(A+A)$, or $\sin(2A)$.

This simple substitution leads to a new set of powerful tools called the double angle identities. They are crucial for simplifying expressions and solving equations, especially in calculus.

For cosine, setting $B=A$ in $\cos(A+B)$ gives us $\cos(2A) = \cos^2 A - \sin^2 A$. But we can do even more. Using the fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1$, we can create two other useful forms of the $\cos(2A)$ formula.

The tangent double angle formula is derived similarly: $\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}$

These identities are the backbone of many 'proving-type' questions you'll see in Section D of your exams. The key is to look at the structure of what you need to prove and decide which identity will transform your expression into the desired form.

For example, to prove that $\frac{\sin 2x}{1 + \cos 2x} = \tan x$, you would start with the left-hand side. Use $\sin 2x = 2\sin x \cos x$ in the numerator. For the denominator, choose the version of $\cos 2x$ that will help you cancel terms. Using $\cos 2x = 2\cos^2 x - 1$ is perfect here, because the '+1' and '-1' will cancel out.

LHS = $\frac{2\sin x \cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x \cos x}{2\cos^2 x} = \frac{\sin x}{\cos x} = \tan x$ = RHS.

Solving an equation like $\sin x = \frac{1}{2}$ has more than one answer. You know that $x=30°$ (or $\frac{\pi}{6}$ radians) is one solution. But because the sine function is periodic, it hits the value $\frac{1}{2}$ again and again, every time it goes around the circle.

We need a way to describe all possible solutions. This is called the general solution

The key is to use the periodicity of the trigonometric functions. For sine, the period is $2\pi$ (or 360°). So if $x$ is a solution, then $x + 2n\pi$ is also a solution for any integer $n$. But there's another angle in each cycle with the same sine value: $\pi - x$.

For cosine and tangent, the general solutions are a bit simpler.

Mastering these identities and solution methods is a matter of practice. It allows you to transform and simplify complex trigonometric expressions, a skill you will rely on heavily in calculus.