Prove that cos^3x - sin^3x/cosx - sinx = 1/2(2 + sin2x)

Question

Prove that

\(\cfrac{cos{^3}x-sin^{3}x}{cosx-sinx}=\frac{1}{2}(2+sin2x)\)

cos³x - sin³x/cosx - sinx = 1/2(2 + sin2x)

Answer 1

To Prove: \(\frac{cos^{3}x-sin^{3}x}{cosx-sinx}=\frac{1}{2}(2+sin2x)\)

Taking LHS,

= \(\frac{cos^{3}x-sin^{3}x}{cosx-sinx}\) .....(i)

We know that,

a³ – b³ = (a – b)(a² + ab + b² )

So, cos³x – sin³x = (cosx – sinx)(cos²x + cosx sinx + sin²x)

So, eq. (i) becomes

= cos²x + cosx sinx + sin2x

= (cos²x + sin²x) + cosx sinx

= (1) + cosx sinx [∵ cos² θ + sin2 θ = 1]

= 1 + cosx sinx

Multiply and Divide by 2, we get

= RHS

∴ LHS = RHS

Hence Proved