Question

Prove the following identities:

`(cos^3θ + sin^3θ)/(cos θ + sin θ) + (cos^3θ - sin^3θ)/(cos θ - sin θ) = 2`

Answer 1

Given: `(cos^3θ + sin^3θ)/(cos θ + sin θ) + (cos^3θ - sin^3θ)/(cos θ - sin θ)` (assume cos θ ± sin θ ≠ 0)

To Prove: `(cos^3θ + sin^3θ)/(cos θ + sin θ) + (cos^3θ - sin^3θ)/(cos θ - sin θ) = 2`

Proof [Step-wise]:

1. Use the sum and difference of cubes factorizations:

a³ + b³ = (a + b)(a² – ab + b²)

a³ − b³ = (a – b)(a² + ab + b²)

2. Apply them with a = cos θ, b = sin θ:

`(cos^3θ + sin^3θ)/(cos θ + sin θ) = cos^2θ - cos θ sin θ + sin^2θ` 

= (cos²θ + sin²θ) – cos θ sin θ   ...(Since cos²θ + sin²θ = 1)

= 1 – cos θ sin θ 

3. Similarly, `(cos^3θ - sin^3θ)/(cos θ - sin θ) = cos^2θ + cos θ sin θ + sin^2θ` 

= (cos²θ + sin²θ) + cos θ sin θ 

= 1 + cos θ sin θ

4. Add the two results:

(1 – cos θ sin θ) + (1 + cos θ sin θ)

= 1 + 1 

= 2

Therefore `(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ − sin^3 θ)/(cos θ − sin θ) = 2`, as required for angles where cos θ ± sin θ ≠ 0.