To maximise a trigonometric function

[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?

 

[Cyosis]

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]

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
1^2+ (\sqrt{3})^2}= 1+ 3= 4[/itex]<br /> Okay, write that as <br /> 2(\frac{1}{2}sin(x)+(-\frac{\sqrt{3}}{2}cos(x))[/itex] &lt;br /&gt; &lt;br /&gt; 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&amp;#039;s easy to find, but it really doesn&amp;#039;t matter what \theta is.

 

[soopo]

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
1^2+ (\sqrt{3})^2}= 1+ 3= 4[/itex]<br /> Okay, write that as <br /> 2(\frac{1}{2}sin(x)+(-\frac{\sqrt{3}}{2}cos(x))[/itex] &lt;br /&gt; &lt;br /&gt; 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&amp;#039;s easy to find, but it really doesn&amp;#039;t matter what \theta is.

&lt;br /&gt; &lt;br /&gt; Thank you for your answer!&lt;br /&gt; &lt;br /&gt; I really remember that to raise to the power 2 is often useful in trig calculations.&lt;br /&gt; It seems that this time the strategy:&lt;br /&gt; 1. raise to the power 2&lt;br /&gt; 2. and consider the extreme situation&lt;br /&gt; did not work.&lt;br /&gt; &lt;br /&gt; I believe that you can get the same result also with my strategy.&lt;br /&gt; However, I am unsure how to get 2 out of my formula.&lt;br /&gt; The best which I can do is&lt;br /&gt; 3^.5