# Thread: max value of trigonometric function

1. ## max value of trigonometric function

Hi, can someone please explain how to solve the problem below.

Find the maximum value of 3sinx + cos2x.

Your help is very much appreciated.

2. ## Re: max value of trigonometric function

There are a few ways. If you know calculus, take the derivative and find the critical points.

If you don't know calculus. You can also graph the equation either by hand with several test test points or with a graphing utilitiy.

If you prefer to not graph it, you can do some critical thinking. For example, on the interval from x=0 to x = pi/2, you have 3sinx going from 0 to 3. On this same interval you have cos2x going from 1 to -1. Thus you can determine which one changes value faster. Then use that information to figure out at what angle the fastest growing function stop growing.

3. ## Re: max value of trigonometric function

Originally Posted by shiny718
Hi, can someone please explain how to solve the problem below.

Find the maximum value of 3sinx + cos2x.

Your help is very much appreciated.
$y = 3\sin{x} + \cos(2x)$

$y = 3\sin{x} + 1 - 2\sin^2{x}$

$y = -2\sin^2{x} + 3\sin{x} + 1$

y is quadratic in $sin{x}$ ...

maximum value of y occurs at $\sin{x} = \left(\frac{-b}{2a}\right)$ ... $\sin{x} = \frac{3}{4}$

$y_{max} = 3\left(\frac{3}{4}\right) + 1 - 2\left(\frac{3}{4}\right)^2 = \frac{17}{8}$

the max value can also be found using the 1st and 2nd derivative tests taught in calculus

4. ## Re: max value of trigonometric function

If that were "3 cos(x)+ 2sin(x)" you could use r sin(x+ a)= r sin(a)cos(x)+ r cos(a)sin(x) and look for a and r so that r cos(a)= 3, r, sin(a)= 2 and $r^2cos^2(a)+ r^2 sin^2(a)= r^2= 3^2+ 2^2$. (I say this because I misread the problem and started to do it that way!)

But with "3 cos(x)+ sin(2x)" I see no method simpler than taking the first derivative and setting it equal to 0- a "Calculus" rather than "Trigonometry" method.

5. ## Re: max value of trigonometric function

Originally Posted by HallsofIvy
If that were "3 cos(x)+ 2sin(x)" you could use r sin(x+ a)= r sin(a)cos(x)+ r cos(a)sin(x) and look for a and r so that r cos(a)= 3, r, sin(a)= 2 and $r^2cos^2(a)+ r^2 sin^2(a)= r^2= 3^2+ 2^2$. (I say this because I misread the problem and started to do it that way!)

But with "3 cos(x)+ sin(2x)" I see no method simpler than taking the first derivative and setting it equal to 0- a "Calculus" rather than "Trigonometry" method.
Except that the function is \displaystyle \begin{align*} 3\sin{x} + \cos{2x} \end{align*}, and can be reduced to a quadratic where the maximum can be found as shown in Skeeter's post.

6. ## Re: max value of trigonometric function

Thanks a lot!