(cos³A + sin³A)/(cosA + sinA) = 1/2 (2 - sin2A) | Prove | Trigonometry

Prove: $\dfraf{cos³A + sin³A}{cosA + sinA} = \frac{1}{2} (2 - sin2A)$

Solution:

LHS

$= \dfrac{cos³A + sin³A}{cosA + sinA}$

$= \dfrac{(cosA)^3 + (sinA)^3}{cosA + sinA}$

Expand the numerator using the factorisation formula of (a³+b³)

$= \dfrac{(cosA + sinA)(cos²A - sinAcosA + sin²A)}{(cosA + sinA)}$

$= cos²A - sinAcosA + sin²A$

$= (cos²A + sin²A) - sinAcosA$

Use the identity of (cos²A + sin²A = 1)

$= (1- sinAcosA)$

Multiply and divide the expression by 2 to get closer to your RHS

$= \dfrac{2}{2} × (1 - sinAcosA)$

$= \dfrac{1}{2} (2-2sinAcosA)$

Use the formula of sin2A to contract the expression

$= \dfrac{1}{2} (2-  sin2A)$

RHS

See more solutions of Trigonometry:

1. Multiple Angles
2. Sub Multiple Angles