Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
Factor the numerator - it's the difference of cubes formula. Then use the fundamental identity – The Chaz 2.0 Jan 5 at 14:33
    
The identity is false for alpha = pi/4. – djechlin Jan 5 at 15:48
4  
@djechlin An "identity" in this context certainly means that the functions are identical on their common domain. – Steven Gubkin Jan 5 at 16:08
    
@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. – djechlin Jan 5 at 16:31
1  
@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. – Steven Gubkin Jan 5 at 16:47
up vote 12 down vote accepted

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

share|cite|improve this answer

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

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

share|cite|improve this answer

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

The left hand side and right hand side are identical

share|cite|improve this answer
1  
And assuming $\alpha \neq \pi/4$ or else you multiplied by zero on both sides and claimed it was reversible. – djechlin Jan 5 at 15:49
    
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. – Teepeemm Jan 5 at 23:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.