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