# To maximise a trigonometric function

## Homework Statement

What is the maximal value of
f(x) = sin(x) - $$\sqrt{3} cos(x)$$
such that
0 <= x < 360?

## The Attempt at a Solution

To raise to power to two

$$f^2 = sin^2(x) - 2 \sqrt{3} sinx cosx + 3cos^2(x)$$

I obverse that the last term has the greatest coeffient so
I maximise it by
setting x = 0.

We get
f^2(0) = 3
and

f = 3^.5

However, the right answer is 2.

How can you get the right answer?

Cyosis
Homework Helper
Why did you take the square? To maximize the function you need to take the derivative of f with respect to x and equal it to zero. You will find two types of x-es, one that minimizes and one that maximizes. Start with calculating f'(x).

HallsofIvy
Homework Helper
Another way to do that, without differentiating, is to use the trig identity sin(r+ t)= sin(r)cos(t)+ cos(r)sin(t). Of course, you can't find t such that sin(t)= 1 and $cos(t)= \sqrt{3}$ because you must have $cos^2(t)+ sin^2(t)= 1$ while
[tex]1^2+ (\sqrt{3})^2}= 1+ 3= 4[/itex]
Okay, write that as
[tex]2(\frac{1}{2}sin(x)+(-\frac{\sqrt{3}}{2}cos(x))[/itex]

Then $sin(x)+ \sqrt{3}cos(x)= 2 sin(x+ \theta)$ where $\theta$ is the angle such that $cos(\theta)= 1/2$ and $cos(\theta)= \sqrt{3}/2$. That's easy to find, but it really doesn't matter what $\theta$ is.

Another way to do that, without differentiating, is to use the trig identity sin(r+ t)= sin(r)cos(t)+ cos(r)sin(t). Of course, you can't find t such that sin(t)= 1 and $cos(t)= \sqrt{3}$ because you must have $cos^2(t)+ sin^2(t)= 1$ while
[tex]1^2+ (\sqrt{3})^2}= 1+ 3= 4[/itex]
Okay, write that as
[tex]2(\frac{1}{2}sin(x)+(-\frac{\sqrt{3}}{2}cos(x))[/itex]

Then $sin(x)+ \sqrt{3}cos(x)= 2 sin(x+ \theta)$ where $\theta$ is the angle such that $cos(\theta)= 1/2$ and $cos(\theta)= \sqrt{3}/2$. That's easy to find, but it really doesn't matter what $\theta$ is.