# max value of trigonometric function

• Aug 2nd 2012, 06:28 AM
shiny718
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.
• Aug 2nd 2012, 07:04 AM
Dagger2006
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.
• Aug 2nd 2012, 07:04 AM
skeeter
Re: max value of trigonometric function
Quote:

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.

$\displaystyle y = 3\sin{x} + \cos(2x)$

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

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

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

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

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

the max value can also be found using the 1st and 2nd derivative tests taught in calculus
• Aug 2nd 2012, 07:16 AM
HallsofIvy
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 $\displaystyle r^2cos^2(a)+ r^2 sin^2(a)= r^2= 3^2+ 2^2$. (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.
• Aug 2nd 2012, 07:21 AM
Prove It
Re: max value of trigonometric function
Quote:

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 $\displaystyle r^2cos^2(a)+ r^2 sin^2(a)= r^2= 3^2+ 2^2$. (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 \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.
• Aug 3rd 2012, 05:15 AM
shiny718
Re: max value of trigonometric function
Thanks a lot! :)