verify the identity:
(sin^3x-cos^3x)/(sinx-cosx)=1+sinxcosx
the left side top factors using the difference of cubes
LS = (sinx - cosx)(sin^2x + sinxcosx + cos^2x)/sinx-cosx)
= 1 + sinxcosx
= RS
To verify the given identity, we need to simplify both sides of the equation and then check if they are equal.
Let's start by simplifying the left-hand side (LHS) of the equation:
(sin^3x - cos^3x) / (sinx - cosx)
We can factor the numerator using the difference of cubes formula, which states that a^3 - b^3 = (a - b)(a^2 + ab + b^2):
(sin^3x - cos^3x) = (sinx - cosx)(sin^2x + sinxcosx + cos^2x)
Now, the LHS becomes:
[(sinx - cosx)(sin^2x + sinxcosx + cos^2x)] / (sinx - cosx)
Next, we can cancel out the common factor of (sinx - cosx):
sin^2x + sinxcosx + cos^2x
Using the Pythagorean identity sin^2x + cos^2x = 1, we can replace sin^2x + cos^2x with 1:
1 + sinxcosx
Therefore, we have simplified the LHS to 1 + sinxcosx, which is the same as the right-hand side (RHS) of the equation.
Hence, we have verified the identity:
(sin^3x - cos^3x) / (sinx - cosx) = 1 + sinxcosx