Question

Simplify the expression (sin^3x+cos^3x)/(sinx+cosx)

Answer 1

the sum of two cubes \(a^3+b^3 \) factors as \[(a+b)(a^2-ab+b^2)\]

Answer 2

making \[\frac{a^3+b^3}{a+b}=a^2-ab+b^2\] after factoring and cancelling

Answer 3

thanks

Answer 4

yw

Answer 5

wait

Answer 6

To simplify this expression, Am I looking for cos(theta +beta)?

Answer 7

cos(4x+3x)= cos(4x)cos(3x)- sin(4x)sin(3x)
It would be nice. Are we sure of the question?

Answer 8

and cos(4x + 3x) = cos(4x)cos(3x) - sin(4x)sin(3x) and not the question asked

Answer 9

The simplest I see is to factor out cos(3x)
cos(3x)(cos(4x)-sin(4x)

Answer 10

@Ajay1010 Let \(a = \sin x\) and \(b = \cos x\)

then you can rewrite the original problem as

\[\frac{\sin^3x + \cos^3x}{\sin x + \cos x} = \frac{a^3+b^3}{a+b} = \frac{\cancel{(a+b)}(a^2-ab+b^2)}{\cancel{a+b}} = a^2-ab+b^2\]Now undo the substitution we made:
\[\frac{\sin^3 x + \cos^3 x}{\sin x + \cos x}=(\sin x)^2 - (\sin x)(\cos x) + (\cos x)^2 \]\[=\sin^2 x-\sin x\cos x + \cos^2 x = \sin^2 x + \cos^2 x - \sin x\cos x\]
but \(\sin^2 x + \cos^2 x = 1\) so we can further simplify to
\[\frac{\sin^3 x + \cos^3 x}{\sin x + \cos x}= 1-\sin x\cos x\]

@silveralchemist09 You have misread the problem