# To maximise a trigonometric function

• soopo
In summary, the maximal value of the function f(x) = sin(x) - \sqrt{3} cos(x) such that 0 <= x < 360 is 2. This can be found by taking the derivative of f with respect to x and setting it equal to zero, or by using the trig identity sin(x+ t)= sin(x)cos(t)+ cos(x)sin(t) and finding the angle \theta such that cos(\theta)= 1/2 and cos(\theta)= \sqrt{3}/2. The strategy of raising to the power of two and considering the extreme situation did not work in this case.
soopo

## 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?

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).

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.

HallsofIvy said:
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.

I really remember that to raise to the power 2 is often useful in trig calculations.
It seems that this time the strategy:
1. raise to the power 2
2. and consider the extreme situation
did not work.

I believe that you can get the same result also with my strategy.
However, I am unsure how to get 2 out of my formula.
The best which I can do is
3^.5

## 1. What is the definition of a trigonometric function?

A trigonometric function is a mathematical function that relates the angles of a right triangle to the lengths of its sides. The most commonly used trigonometric functions are sine, cosine, and tangent.

## 2. How do you maximize a trigonometric function?

To maximize a trigonometric function, you need to find the maximum value of the function within a specific interval. This can be done by taking the derivative of the function, setting it equal to zero, and solving for the critical points. Then, plug these critical points into the original function to find the maximum value.

## 3. Why is it important to maximize a trigonometric function?

Maximizing a trigonometric function can be useful in many real-world applications, such as finding the maximum height of a projectile or the maximum profit in a business scenario. It allows us to optimize our solutions and make the most efficient use of our resources.

## 4. What are some common techniques for maximizing trigonometric functions?

Some common techniques for maximizing trigonometric functions include using the first or second derivative test, finding critical points, and using trigonometric identities to simplify the function. Graphing the function can also provide insight into the maximum value.

## 5. Can any trigonometric function be maximized?

Yes, any trigonometric function can be maximized within a specific interval. However, some functions may not have a maximum value, such as tangent or cosecant, as they continue to increase or decrease without bound.

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