4
$\begingroup$

I want to verify the following identities:

$${\sin^3\alpha-\cos^3\alpha\over \sin\alpha -\cos\alpha} = 1 + \sin\alpha \cos\alpha$$

I feel like I need to work on the first member – the second one looks fine. I can't really figure out how to transform the first one. Any hints?

$\endgroup$
8
  • $\begingroup$ Factor the numerator - it's the difference of cubes formula. Then use the fundamental identity $\endgroup$ Jan 5, 2016 at 14:33
  • $\begingroup$ The identity is false for alpha = pi/4. $\endgroup$
    – djechlin
    Jan 5, 2016 at 15:48
  • 4
    $\begingroup$ @djechlin An "identity" in this context certainly means that the functions are identical on their common domain. $\endgroup$ Jan 5, 2016 at 16:08
  • $\begingroup$ @StevenGubkin really depends, you can't really leave things like that "implied" until mastery is assumed (both college courses and research papers seem to operate on this principle). Given that this problem looks precalc-level seems reasonable to point out the necessary hypothesis. $\endgroup$
    – djechlin
    Jan 5, 2016 at 16:31
  • 1
    $\begingroup$ @djechlin I certainly think that pointing out the domain issue is a good thing to do, but I do not think that it invalidates the word "identity" in this situation. Is $\sqrt{x^2} = |x|$ not an identity unless we specify that it is over the reals? I think the assumed context is acceptable if the course deals entirely with real numbers. $\endgroup$ Jan 5, 2016 at 16:47

3 Answers 3

12
$\begingroup$

Hint: $u^3-v^3 = (u-v) (u^2+u v+v^2)$

$\endgroup$
3
$\begingroup$

Assuming $x \neq y$, $$\dfrac{x^3 - y^3}{x-y} = x^2 + xy + y^2$$

And in your case, $x^2 + y^2 = 1$.

$\endgroup$
3
$\begingroup$

$$\frac{\sin^3(a)-\cos^3(a)}{\sin(a)-\cos(a)}=1+\sin(a)\cos(a)\Longleftrightarrow$$ $$\cos^2(a)+\cos(a)\sin(a)+\sin^2(a)=1+\sin(a)\cos(a)\Longleftrightarrow$$ $$\cos^2(a)+\sin^2(a)=1\Longleftrightarrow$$ $$1=1$$

The left hand side and right hand side are identical

$\endgroup$
2
  • 1
    $\begingroup$ And assuming $\alpha \neq \pi/4$ or else you multiplied by zero on both sides and claimed it was reversible. $\endgroup$
    – djechlin
    Jan 5, 2016 at 15:49
  • $\begingroup$ Going from the first step to the second needs a comment as to what's going on. Whoever can see what happened wouldn't have needed to ask how to prove the identity in the first place. $\endgroup$
    – Teepeemm
    Jan 5, 2016 at 23:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .