<Dd> c 2 sin 2 ⁡ α (1 + c 2 sin 2 ⁡ α (b − c cos ⁡ α) 2) = a 2 ⋅ c 2 sin 2 ⁡ α (b − c cos ⁡ α) 2 (\ displaystyle c ^ (2) \ sin ^ (2) \ alpha \ left (1 + (\ frac (c ^ (2) \ sin ^ (2) \ alpha) ((b-c \ cos \ alpha) ^ (2))) \ right) = a ^ (2) \ cdot (\ frac (c ^ (2) \ sin ^ (2) \ alpha) ((b-c \ cos \ alpha) ^ (2)))) </Dd> <P> By multiplying by (b − c cos α), we can obtain the following equation: </P> <Dl> <Dd> (b − c cos ⁡ α) 2 + c 2 sin 2 ⁡ α = a 2 . (\ displaystyle (b-c \ cos \ alpha) ^ (2) + c ^ (2) \ sin ^ (2) \ alpha = a ^ (2).) </Dd> </Dl> <Dd> (b − c cos ⁡ α) 2 + c 2 sin 2 ⁡ α = a 2 . (\ displaystyle (b-c \ cos \ alpha) ^ (2) + c ^ (2) \ sin ^ (2) \ alpha = a ^ (2).) </Dd>

The law of cosines reduces to the pythagorean theorem when the triangle is acute