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## Homework Statement

What is the maximal value of

f(x) = sin(x) - [tex]\sqrt{3} cos(x)[/tex]

such that

0 <= x < 360?

## The Attempt at a Solution

To raise to power to two

[tex] f^2 = sin^2(x) - 2 \sqrt{3} sinx cosx + 3cos^2(x) [/tex]

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?