You already know how to find the sine of 30 degrees and that $\sin^2 x + \cos^2 x = 1$. Now, we move into the more strategic part of trigonometry. Think of it less like calculation and more like solving a puzzle. The goal is to prove that one complex expression is actually equal to another, often simpler, one. Mastering this requires a new set of tools for transforming expressions.

One of the most powerful techniques is changing sums of trigonometric functions into products. This often simplifies expressions dramatically, especially when you need to cancel terms. These transformations are commonly known as the C and D formulas because of the variables often used to represent the angles.

Here are the key formulas for converting sums and differences into products:

Let's see this in action. Suppose you need to prove the identity:

$\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

On their own, the terms on the left don't simplify. But by applying the sum-to-product formulas, we can transform the numerator and the denominator.

Simplifying the angles inside the functions gives us:

After cancelling, we're left with $\frac{\sin 4x}{\cos 4x}$, which is simply $\tan 4x$. The identity is proven. The key was transforming the sums into products to reveal common factors.

Some identities are only true under specific conditions. A very common condition in problems is that the angles of a triangle, $A$, $B$, and $C$, are involved. This gives us a powerful hidden tool: the fact that $A + B + C = \pi$ radians (or 180°). This constraint allows for clever substitutions. Conditional identities often require you to isolate a variable, like $A+B = \pi - C$, and then apply trigonometric properties.

Let's prove that if $A+B+C=\pi$, then $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$.

This looks intimidating. We start by working on the left-hand side (LHS), grouping the first two terms and applying the sum-to-product formula:

Now, we use our condition. Since $A+B+C=\pi$, we know that $A+B = \pi - C$. This means $\sin(A+B) = \sin(\pi - C) = \sin C$. Let's substitute that in.

We can factor out $2\sin C$:

Let's use the condition again. We can write $\cos C = \cos(\pi - (A+B)) = -\cos(A+B)$. Substituting this gives:

The expression in the parenthesis, $\cos(A-B) - \cos(A+B)$, simplifies to $2\sin A \sin B$. (You can prove this using the sum/difference formulas for cosine or the C and D formulas). Our final expression becomes:

Sometimes, proving an identity seems impossible using standard formulas. A powerful technique for these situations is to express trigonometric functions in terms of the tangent of a half-angle. These half-angle substitutions can convert a complex trigonometric expression into a more straightforward algebraic one.

The core substitutions, where $t = \tan(x/2)$, are:

Consider proving the identity: $\frac{1-\cos x + \sin x}{1+\cos x + \sin x} = \tan\left(\frac{x}{2}\right)$.

Let's substitute the half-angle formulas into the left side, using $t = \tan(x/2)$:

To simplify this complex fraction, we can multiply the numerator and denominator by the common denominator, $1+t^2$.

Now, simplify the algebra in the numerator and denominator:

Finally, we factor and cancel terms:

Since we defined $t = \tan(x/2)$, we have proven that the left side simplifies to $t$, which equals the right side. This technique converted a tricky trigonometric proof into a standard algebraic simplification.

These strategies—transforming sums, using conditions, and substituting—are the core of advanced trigonometric proofs. Practice is key to recognizing which tool to use for each problem.