# Find the maximum value of this expression

1. Jun 12, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
$\dfrac{tan\left( \theta + 2\pi /3 \right) - tan\left( \theta + \pi /6 \right) + cos\left( \theta + \pi /6 \right)}{\sqrt{3}} \\ \theta \in \left( -5 \pi /12, - \pi /3 \right)$

3. The attempt at a solution
At maximum, f'(x)=0

sin(θ+∏/6)=0

But there is no theta in the given interval that satisfies this equation.

2. Jun 12, 2013

### Fightfish

I haven't checked whether you did your differentiation correctly (something that you definitely want to check), but in general, for functions defined on a finite domain, the global maximum and minimum points need not occur at stationary points (which correspond to local minima / maxima). The global maximum and minimum can also be attained at the endpoints of the interval.

3. Jun 12, 2013

### CAF123

To maximise this expression means you want to maximise the numerator of that expression.
Consider each term individually and find θ in domain such that each term is ether maximised or minimised depending on the sign preceding it.

Sketch a graph or note that tan and cos are strictly increasing functions on this domain.

4. Jun 12, 2013

### utkarshakash

5. Jun 12, 2013

### Saitama

Instead of graph, I think you can do it algebraically too.

Rewrite $\tan(\theta+2\pi/3)=-\cot(\theta+\pi/6)$.
Let $\theta+\pi/6=t$. Then $-\pi/4<t<-\pi/6$.

Now it's much easier to solve. You can find the derivative or analyze the graph of the resulting expression.

Hope that helped.

6. Jun 12, 2013

### Ray Vickson

Not on an open interval. On an open interval there can be a supremum at an endpoint, but no maximum. Of course, one might speak informally of a maximum in such a case, but it is not technically correct.