# Homework Help: Limit of trigonometry

1. Feb 17, 2014

### Robertoalva

1. The problem statement, all variables and given/known data
Find the limit

2. Relevant equations
lim θ→ ∏/6 (sinθ - 1/2)/(θ - ∏/6)

3. The attempt at a solution

well, if you substitute the values, it gives 0/0, but that means that is undefined and that's not the answer. according to my book the answer is √(3)/2 and I am not supposed to use L'Hopitals rule. this is we're I need help. Thanks

2. Feb 17, 2014

### Staff: Mentor

Use a substitution (u = θ - $\pi$/6) to rewrite your limit as u → 0. Try to get it into the form $\lim_{u \to 0}\frac{sin(u)}{u}$, which is a well-known limit.

3. Feb 17, 2014

### lurflurf

That is the derivative of sin(x) at x=∏/6. If you know the derivative of sine you are done. The derivative of sine can be expressed in terms of the derivative of sine when x=0. The derivative of sine at x=0 is usually an axiom or easily follows from the axioms.

4. Feb 17, 2014

### Robertoalva

so basically the whole thing gives me 1? or the -1/2 stays and subtracts after i do the limit?

5. Feb 17, 2014

### Staff: Mentor

Make the substitution and see what you get.

6. Feb 17, 2014

### Staff: Mentor

I'm guessing that the aim of the problem is to get the poster to evaluate the limit, but not in the context that it's the derivative of anything.

7. Feb 17, 2014

### Robertoalva

what i have right now is lim u->0 (sin(u) - (1/2))/u. after that I know that lim u->0 sin(u)/u= 1 then 1-(1/2). but it gives me 1/2 and that's not the correct answer. what am I doing wrong?

8. Feb 17, 2014

### Robertoalva

??????

9. Feb 17, 2014

### SammyS

Staff Emeritus
That's not what you should be getting.

Show how you get that.

10. Feb 17, 2014

### Robertoalva

that's how I did it... sin(u)/u= 1 so 1-(1/2)=1/2

11. Feb 17, 2014

### Mentallic

What's the expression you ended up with to conclude this? Show us each step you've taken, starting with the u substitution.

12. Feb 17, 2014

### SammyS

Staff Emeritus
Mark 44 suggested that you use the substitution,
$\displaystyle u=\theta-(\pi/2) \$ .​

Of course, that gives
$\displaystyle \theta=u+(\pi/2)$

You need to use the angle addition formula for
$\displaystyle \sin\left(u+(\pi/2)\right)$​

13. Feb 20, 2014

### Robertoalva

but he said that u= theta - (pi/6) not pi/2

14. Feb 20, 2014

### Robertoalva

according to what i have read, sin(u+(pi/2))= cos u but, can I do it with sin(u+(pi/6)) ?

15. Feb 20, 2014

### Mentallic

No, you can't, but do you know how to expand sin(A+B)?

16. Feb 20, 2014

### Robertoalva

yes sin(u+v)=sin(u)cos(v)+cos(u)sin(v) in which this case would be sin(u+(pi/2))= sin(u)cos(pi/2)+cos(u)sin(pi/2)

17. Feb 20, 2014

### SammyS

Staff Emeritus
Yup, Mentallic is right.

Ok, two problems. We're supposed to be making the substitution $u = \theta - \pi/6$ instead, and expressions like $\cos(\pi/2)$ can be simplified.