Question

(sin A + cos A) = (1 — sinA cosA) = sin³ A + cos³A prove it.​

Answer 1

Answer:

How do I prove (sinA - cosA) (1-sinAcosA) = sin^3A+cos^3A where A is the size of an angle?

It might not be true. Try A=45º . The left side is 0 but the right side is not.

HOWEVER

sin3A+cos3A=(sinA+cosA)(sin2A−sinAcosA+cos2A)=(sinA+cosA)(sin2A+cos2A−sinAcosA)=(sinA+cosA)(1−sinAcosA)

(Actually, there is a mistake in the statement. The negative sign in the first parenthesis should be a positive. With this remedied, proceed as follows.)

The principal trick, here is to recall that 1 = sin^2 A + cos^2 A .

The other term in the parentheses, being second degree in sin A and cos A , suggests that that might be useful.

Alternatively, start with the RHS, a difference of two cubes, which has a standard factorisation:

a^3 + b^3 = (a + b) (a^2 - ab + b^2) .

So sin^3 A + cos^3 A = (sin A + cos A) (sin^2 A - sin A cos A + cos^2 A)

= (sin A + cos A) (1 - sin A cos A) ,

as required.