Question

How to find the maximum value of f(x) = sin(x) - 2 cos(3x)

Answer 1

Do you know how to find f'(x)?

Answer 2

Yup.
f' = cos(x) + 6sin(3x)

Answer 3

Set f'(x)=0 and solve for x=\(x_{max}\)
Then maximum value of f(x) is
f(\(x_{max}\)).
Notice that \(x_{max}\) has multiple roots, because sine and cosine are periodic functions.

Answer 4

cos(x) + 6 sin(3x) = 0
cos(x) + 6(sin(x+2x)) = 0
cos(x) + 6(sin(x)cos(2x) + cos(x)sin(2x)) = 0
cos(x) + 6sin(x)(2cos^2 x - 1) + 12sin(x)cos^2 x) = 0
cos(x) - 6sin(x) + 24sin(x)cos^2 x = 0
So far i was stuck ...

Answer 5

Try simplifying using the identity:
\(sin(3x)=3cos^2(x)sin(x)-sin^3(x)\)
and go from there, unless there is a shorter way.

Answer 6

and then ... ?
Actually i hope the equation become into sin only or just cos only. Hmmm still confuse to solve for x

Answer 7

Yeah, it comes out to be a quartic with both sine and cosine, and odd powers for both.
I'll have to think about it, or you can tag someone else in the mean time.

Answer 8

* cubic
Need to solve:
\(f'(x)=-6sin^3(x)+18cos^2(x)sin(x)+cos(x)=0\)

Answer 9

Numerical solution is easy with Newtons, but with three relative maxima from 0 to 2pi.

Answer 10

yeah... looks the equation cant be solved just use basic trigono, right ?

Answer 11

It should be possible, just need a good approach! lol

Answer 12

.

Answer 13

we can try to rewrite the function, using the subsequent identity:
\[\cos \left( {3x} \right) = {\left( {\cos x} \right)^3} - {\left( {\sin x} \right)^2}\cos x\]

Answer 14

or we can try to go in the complex, namely:
\[f\left( x \right) = \frac{{ - i}}{2}\left( {{e^{ix}} - {e^{ - ix}}} \right) - {e^{i3x}} - {e^{ - i3x}}\]

Answer 15

or,

Well, it's solvable, but reduces to the solution of a polynomial in cubic.
This is what I did.
We're solving
f'(x)=-6sin(x)^3+18cos(x)^2sin(x)+cos(x)=0
By substitution, we know that sin(0), sin(pi) and multiples are not solutions.
So I factor out \(sin^3(x)\) to get
\(f'(x)=sin^3(x)(-6+18cot^2(x)+cot(x)(1+cot^2(x))\)... using 1+cot^2(x)=csc(x)
So let y=cot(x), and then solve the algebraic equation
g(y)=y^3+18y^2+y-6=0
where
y={[0.5425152410243347,-17.92554099585428,-0.6169742451700351]}
However, some manipulations with pi and 2pi are required to get the proper values of x, where the maxima are at the following locations:
{1.07372,3.08586,5.265196}

Answer 16

using my last substitution, I got this:
\[f'\left( x \right) = 0\quad \Rightarrow {e^{2ix}}\left( {\frac{i}{2} + 3} \right) - \left( {\frac{i}{2} - 3} \right) = 0\]

Answer 17

I'm very sorry, I have made an error, here is the right expression:
\[f'\left( x \right) = 0\quad \Rightarrow {e^{2ix}}\left( {\frac{i}{2} + 3{e^{i2x}}} \right) + \left( {\frac{i}{2} - 3{e^{i2x}}} \right) = 0\]

Answer 18

I think that we can use this parametrization:
\[\begin{gathered}
t = \tan \left( {\frac{x}{2}} \right) \hfill \\
\hfill \\
\sin x = \frac{{2t}}{{1 + {t^2}}},\quad \cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}} \hfill \\
\end{gathered} \]

Answer 19

with the above parametrization, I get this:
\[\large f\left( x \right) = 2\frac{{{t^6} + {t^5} - 15{t^4} + 2{t^3} + 15{t^2} + t - 1}}{{{{\left( {1 + {t^2}} \right)}^3}}}\]

Answer 20

and here is the first derivative:
\[\large \frac{{ - {t^6} + 6{t^5} - {t^4} - 120{t^3} + {t^2} + 36t + 1}}{{{{\left( {1 + {t^2}} \right)}^4}}}\]