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±B) and cos(A±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=2C+D,B=2C−D) 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∘ 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 ±60∘ around it (e.g. 10∘,30∘-adjacent-triples like 10∘,50∘,70∘, or 12∘,48∘-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=π). 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=π−A−B, so cosC=−cos(A+B), sinC=sin(A+B), or at the half-angle level sin2C=cos2A+B), apply a sum-to-product step, and repeat until a single compact product remains. This is exactly how the standard triangle identities are built:
together with the bound 1<cosA+cosB+cosC≤23 that follows from the first identity plus the fact sin2Asin2Bsin2C≤81 (itself proved by treating that product as a quadratic in cos2A−B and demanding a non-negative discriminant).
Worked illustration (mixing both directions). To show cos36∘cos72∘cos108∘cos144∘=161: rewrite cos108∘=cos(90∘+18∘)=−sin18∘ and cos144∘=cos(180∘−36∘)=−cos36∘, and cos72∘=sin18∘, so the product becomes cos36∘⋅sin18∘⋅(−sin18∘)⋅(−cos36∘)=sin218∘cos236∘. Substituting the standard surd values sin18∘=45−1 and cos36∘=45+1 gives (45−1)2(45+1)2=[16(5−1)(5+1)]2=[164]2=[41]2=161. No sum-to-product step was even needed here beyond angle relations — the point is that recognising which identity or angle relation applies (product-to-sum, sum-to-product, complementary/supplementary angle, or a known surd value) is the real skill this concept builds, well beyond memorising the eight boxed formulas.
Use A+B=π−C (so sin(A+B)=sinC,cos(A+B)=−cosC) throughout; each part reduces to a sum-to-product step followed by this substitution.
✓Final answer
All seven identities hold whenever A+B+C=180∘; see the worked steps for each.
Every part uses the same substitution A+B=180∘−C (so sin(A+B)=sinC and cos(A+B)=−cosC), combined with a sum-to-product or double-angle step.
Step 1. Part (i): sin2A+sin2B+sin2C=4sinAsinBsinC.sin2A+sin2B=2sin(A+B)cos(A−B)=2sinCcos(A−B) (using sin(A+B)=sinC). Also sin2C=2sinCcosC. Adding: 2sinC[cos(A−B)+cosC]. Since cosC=−cos(A+B): cos(A−B)+cosC=cos(A−B)−cos(A+B)=2sinAsinB. So the sum is 2sinC⋅2sinAsinB=4sinAsinBsinC.
Step 2. Part (ii): cosA+cosB−cosC=−1+4cos2Acos2Bsin2C.cosA+cosB=2cos2A+Bcos2A−B=2sin2Ccos2A−B (since 2A+B=90∘−2C). And −cosC=−1+2sin22C. Adding: −1+2sin2C[cos2A−B+sin2C]. Since sin2C=cos2A+B: cos2A−B+cos2A+B=2cos2Acos2B. So the total is −1+2sin2C⋅2cos2Acos2B=−1+4cos2Acos2Bsin2C.
Step 3. Part (iii): sin2A+sin2B+sin2C=2+2cosAcosBcosC. Using sin2θ=1−cos2θ and the already-established identity cos2A+cos2B+cos2C=1−2cosAcosBcosC (proved in §3.5.4): sin2A+sin2B+sin2C=3−(cos2A+cos2B+cos2C)=3−(1−2cosAcosBcosC)=2+2cosAcosBcosC.
Step 4. Part (iv): sin2A+sin2B−sin2C=2sinAsinBcosC.sin2A+sin2B=1−21(cos2A+cos2B)=1−cos(A+B)cos(A−B)=1+cosCcos(A−B) (since cos(A+B)=−cosC). Subtract sin2C=1−cos2C: sin2A+sin2B−sin2C=cosCcos(A−B)+cos2C=cosC[cos(A−B)+cosC]. As in Step 1, cos(A−B)+cosC=2sinAsinB, so this is 2sinAsinBcosC.
Step 5. Part (v): tan2Atan2B+tan2Btan2C+tan2Ctan2A=1. Since 2A+B=90∘−2C, tan2A+B=cot2C=1/tan2C. By the tangent addition formula, tan2A+B=1−tan2Atan2Btan2A+tan2B. Equating and cross-multiplying: tan2C(tan2A+tan2B)=1−tan2Atan2B, i.e. tan2Atan2C+tan2Btan2C+tan2Atan2B=1 — exactly the required identity.
Step 6. Part (vi): sinA+sinB+sinC=4cos2Acos2Bcos2C.sinA+sinB=2sin2A+Bcos2A−B=2cos2Ccos2A−B (since sin2A+B=cos2C). And sinC=2sin2Ccos2C. Adding: 2cos2C[cos2A−B+sin2C]. As in Step 2, the bracket is 2cos2Acos2B, so the total is 4cos2Acos2Bcos2C.
Step 7. Part (vii): sin(B+C−A)+sin(C+A−B)+sin(A+B−C)=4sinAsinBsinC. Since B+C=180∘−A: B+C−A=180∘−2A, so sin(B+C−A)=sin(180∘−2A)=sin2A. Likewise C+A−B=180∘−2B⇒sin(⋅)=sin2B, and A+B−C=180∘−2C⇒sin(⋅)=sin2C. So the left side is sin2A+sin2B+sin2C, which equals 4sinAsinBsinC by part (i).
✓Final answer
All seven identities are proved using A+B=180∘−C together with sum-to-product/double-angle steps, as shown above.
Substitute A+B=180°-C, then apply sum-to-product / double-angle repeatedly
Forgetting sin2A+B=cos2C and cos2A+B=sin2C (the half-angle complementary swap)
In part (v), cross-multiplying tan2A+B=cot2C incorrectly and dropping a term