Prove that \(\left(\frac{cos^3 \ A - sin^3 \ A}{cos \ A - sin \ A} \right) - \left(\frac{cos^3 \ A + sin^3 \ A}{cos \ A + sin \ A} \right) = 2 \ sin \ A \ cos \ A\)
Proof:
LHS \(= \left(\frac{cos^3 \ A - sin^3 \ A}{cos \ A - sin \ A} \right) - \left(\frac{cos^3 \ A + sin^3 \ A}{cos \ A + sin \ A} \right)\)
By applying the formula, and simplyfing we get,
By the known trigonometric identity, we get,
\(= 2 \ cos \ A \ sin \ A\)
\(=\) RHS
Answer variants:
\(cos^2 \ A + sin^2 \ A=1\)
\(a^3 - b^3 = (a - b)(a^2 + b^2 + 2ab)\) and \(a^3 + b^3 = (a + b)(a^2 + b^2 - 2ab)\)]
\(cos^2 \ A -sin^2 \ A=1\)
\(= cos^2 \ A + sin^2 \ A + cos \ A \ sin \ A - (cos^2 \ A + sin^2 \ A - cos \ A \ sin \ A)\)
\(a^3 - b^3 = (a - b)(a^2 + b^2 + ab)\) and \(a^3 + b^3 = (a + b)(a^2 + b^2 - ab)\)]
\(a^3 - b^3 = (a + b)(a^2 + b^2 - ab)\) and \(a^3 + b^3 = (a - b)(a^2 + b^2 +ab)\)]