Q.If $\triangle ABC$ is a right triangle and if $\angle A = \dfrac{\pi}{2}$, then prove that

The problem this concept solves. Trigonometric functions naturally arise as products of angle expressions in some settings (e.g. amplitude modulation, or three angles of a triangle multiplied together) and as sums in others (e.g. combining two waves). Converting cleanly between the two forms is one of the most-used trigonometric skills, and this concept covers every tool needed to do it.

1. Product-to-sum (from the addition formulas). Adding/subtracting the four expansions of $\sin(A\pm B)$ and $\cos(A\pm B)$ in pairs isolates a pure product on one side and a sum/difference on the other:

Use these whenever you are handed a product of two sines/cosines and need a sum.

2. Sum-to-product (the reverse substitution). Setting $C=A+B,\ D=A-B$ (so $A=\frac{C+D}2,\ B=\frac{C-D}2$) and substituting back into the four identities above inverts the process:

Use these whenever you are handed a sum or difference of two sines/cosines and need a product — which is usually the move that lets a numerator and denominator share a cancelling factor, or that shows an expression equals zero (a product is zero the moment one factor is).

3. The 60°±A triple-product family. Applying the product-to-sum idea twice in a row to three factors spaced $60^\circ$ apart gives three compact identities:

These are worth recognising on sight: any time three factors in a product are centred on some angle $A$ and spread $\pm60^\circ$ around it (e.g. $10^\circ,30^\circ$-adjacent-triples like $10^\circ,50^\circ,70^\circ$, or $12^\circ,48^\circ$-style pairs alongside a third term), one of these three identities collapses the triple product to a single term in $3A$ immediately.

4. Conditional identities for a triangle ($A+B+C=\pi$). When the three angles are constrained to sum to a fixed value — above all, the interior angles of a triangle — the sum-to-product identities become the engine for proving relations that are otherwise false. The recipe is always: eliminate one angle via the condition (e.g. $C=\pi-A-B$, so $\cos C=-\cos(A+B)$, $\sin C=\sin(A+B)$, or at the half-angle level $\sin\frac C2=\cos\frac{A+B}2$), apply a sum-to-product step, and repeat until a single compact product remains. This is exactly how the standard triangle identities are built:

$\cos A+\cos B+\cos C=1+4\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2, \qquad \sin A+\sin B+\sin C=4\cos\tfrac A2\cos\tfrac B2\cos\tfrac C2,$ …