# Verify $\frac{\sin^3A + \cos^3A}{\sin A + \cos A} = 1 - \sin A\cos A$

How can I verify the following trigonometric identity? $$\frac{\sin^3 A + \cos^3 A}{\sin A+\cos A} = 1-\sin A\cos A.$$

My work so far is \begin{align*} &\frac{\sin\cos(\sin^2+\cos^2)}{\sin+\cos}\\ &\frac{\sin\cos(1)}{\sin+\cos} \end{align*}

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Use the formula for factoring a sum of cubes (what you did was incorrect). –  David Mitra Jul 17 '12 at 22:06
I think we both saw this formatting problem, Arturo. –  ncmathsadist Jul 17 '12 at 22:07

First: don't use $\sin$ and $\cos$ without arguments. For instance, your last formula, it's easy to get confused and think you are computing, inter alia, $\cos(1)$, which you are not.

Second: it seems that you think that $$\sin^3 A + \cos^3A \text{ is equal to } \sin A\cos A(\sin^2A + \cos^2A).$$ They are not equal; you can verify that by actually multiplying out the right hand side; you will see that you get $$\sin^3A\cos A + \cos^3A\sin A\neq \sin^3A + \cos^3A.$$

There are a few algebraic formulas that show up a lot and it is good to know them. When I was in middle school, they were known as "notable products", because they were, well, notable and showed up a lot. One was the square of a binomial, $(a+b)^2 = a^2+2ab+b^2$; one was the "difference of squares" or "conjugate product": $(a+b)(a-b)=a^2-b^2$.

Two others are the difference-of-cubes and the sum-of-cubes: \begin{align*} a^3-b^3 &= (a-b)(a^2+ab+b^2)\\ a^3+b^3 &= (a+b)(a^2-ab+b^2). \end{align*} Using the second one, you can simplify the left hand side, $$\frac{\sin^3 A + \cos^3 A}{\sin A + \cos A}$$ by letting $a=\sin A$ and $b=\cos A$, and proceed from there.

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Oh I see so now I am left with the answer. Thanks for your help it is I got to remember algebra. –  jose Jul 17 '12 at 22:17

$$a^3+b^3 = (a+b)(a^2-ab+b^2).$$ Use that to factor the numerator.

Also, it is not correct to say that $s^3+c^3 = sc(s^2+c^2)$.

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