# Trigonometry Question: Find the max and min.

1. Dec 19, 2009

### Cuisine123

1. The problem statement, all variables and given/known data
Find the maximum and minimum values of 4cos$$\theta$$-3sin$$\theta$$.

2. Relevant equations
I have no idea.

3. The attempt at a solution
I have no idea how to do this question

2. Dec 19, 2009

### Staff: Mentor

Try to draw a plot - even if it will not give you an exact answer, it may give you some hints.

3. Dec 19, 2009

### Staff: Mentor

The trick to these kinds of problems is to write Asin(x) - Bcos(x) as Csin(x - $\theta$). It can then be seen that the maximum value is C and the minimum value is -C (assuming that C > 0).
4cos$\theta$ - 3sin$\theta$
= 5[(4/5)cos$\theta$ - (3/5)sin$\theta$]

Now what you need to do is find an angle $\alpha$ such that sin($\theta$) = 4/5 and cos($\theta$) = 3/5. Then you can use the identity sinAcosB - cosAsinB = sin(A-B).

4. Dec 19, 2009

### jegues

I'm just curious, can you go into more detail how you generated this from,

5. Dec 19, 2009

### ideasrule

Do you mean why it's true, or why Mark wrote it like that? It's easy to prove why it's true: just expand and you get 4cos x - 3 sin x. As for why it's useful, you want an equation of the form cos(a)cos(b)-sin(a)sin(b), because such an equation is equal to cos(a+b). Since cos(a) can't be 4 and sin(a) can't be 3, Mark decided to factor out a five. You can just as well factor out a ten, or a 100.