Question

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

Answer 1

using a^3-b^3 formula and using the sin2x formula

Answer 2

Answer:

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

Solution :

LHS : $\frac{cos^3x - sin^3x}{cosx - sinx}$

Identity : $a^3-b^3=(a-b)(a^2+ab+b^2)$

$\frac{(cos x - sinx)(cos^2x + sin^2x+cos x sin x)}{cosx - sinx}$

$(cos^2x + sin^2x+cos x sin x)$

Identity : $Sin^2x +Cos^2x= 1$

$(1+cos x sin x)$

Identity : $2 sin x cos x = Sin 2x$

$(1+\frac{Sin2x}{2})$

$(\frac{2+Sin2x}{2})$

$\frac{1}{2}(2+Sin2x)$

Hence LHS = RHS