Maximum Value

[mynamesmurph]

In class we are graphing the sum of trig functions, such as y=3cosx+4sinx

To graph them, we choose common x values, and then add them to get points on the graph for the new function. I've noticed on graphing some of the more complex functions that you really need a lot of x values to truly get the shape, even using all the points on the unit circle sometimes isn't enough to really get an idea of the shape. I thought that knowing the maximum value might be helpful sometimes and after class, I asked my teacher if there was a way to know the maximum value. He was in a hurry and I think he said something like add the amplitudes, which sort of made sense at the time, and thinking about it on the way home I thought if I add the amplitudes that will give me a y value for which I can solve x. (I don't actually know how to solve for x for an equation like the one I posted, any help there?) Anyway, I started graphing some trig functions on a graphing calculator and noticed that the maximum is not always the sum of the amplitudes, and often isn't. Did I not hear him correctly? I'm not going to be tested on this, but am curious if there is a way to know this value. Any help?

 

[Deleted member 4993]

In class we are graphing the sum of trig functions, such as y=3cosx+4sinx

To graph them, we choose common x values, and then add them to get points on the graph for the new function. I've noticed on graphing some of the more complex functions that you really need a lot of x values to truly get the shape, even using all the points on the unit circle sometimes isn't enough to really get an idea of the shape. I thought that knowing the maximum value might be helpful sometimes and after class, I asked my teacher if there was a way to know the maximum value. He was in a hurry and I think he said something like add the amplitudes, which sort of made sense at the time, and thinking about it on the way home I thought if I add the amplitudes that will give me a y value for which I can solve x. (I don't actually know how to solve for x for an equation like the one I posted, any help there?) Anyway, I started graphing some trig functions on a graphing calculator and noticed that the maximum is not always the sum of the amplitudes, and often isn't. Did I not hear him correctly? I'm not going to be tested on this, but am curious if there is a way to know this value. Any help?

Let

k = √(3²+4²) = 5

y = k * [3/k * cos(x) + 4/k * sin(x)]

let

sin(Θ) = 3/k and cos(Θ) = 4/k → Θ = 53.13°

then

y = 5 * sin(x + 53.13°)

Now plot it.....

 

[mynamesmurph]

Thank you for your quick response. Would you mind explaining why you're performing each step? I'm not putting it together.

Thanks.

 

[wjm11]

In class we are graphing the sum of trig functions, such as y=3cosx+4sinx

To graph them, we choose common x values, and then add them to get points on the graph for the new function. I've noticed on graphing some of the more complex functions that you really need a lot of x values to truly get the shape, even using all the points on the unit circle sometimes isn't enough to really get an idea of the shape. I thought that knowing the maximum value might be helpful sometimes and after class, I asked my teacher if there was a way to know the maximum value. He was in a hurry and I think he said something like add the amplitudes, which sort of made sense at the time, and thinking about it on the way home I thought if I add the amplitudes that will give me a y value for which I can solve x. (I don't actually know how to solve for x for an equation like the one I posted, any help there?) Anyway, I started graphing some trig functions on a graphing calculator and noticed that the maximum is not always the sum of the amplitudes, and often isn't. Did I not hear him correctly? I'm not going to be tested on this, but am curious if there is a way to know this value. Any help?

What your teacher told you about adding the amplitudes gives you the maximum y value (amplitude) possible, even though the function may not actually reach that value. Let's look at an example: y=cosx+sinx. The separate functions, y=sinx and y=cosx each have a maximum amplitude of 1. If we added them together, we'd have y = sinx + cosx, and we could easily surmise that the maximum possible y value would be y = 1 + 1 = 2. However, this function never reaches 2 (It only reaches a little above 1.4) because the maximum amplitudes of the two separate functions occur at different x values. Make sense?

So adding the amplitudes allows us to pick a good range for plotting a graph, even though the graph may not hit that upper limit.

Hope that helps.

 

[Ishuda]

What Subhotosh Khan was showing was how to turn the sum of sine and cosine trig functions into a single sine or cosine trig function so that you can then look at the amplitude. If you have not learned the sum and difference formulas for the sine and cosine, you couldn't use it but as an example we'll go through how to do this for the problem you gave [a step by step procedure of what was posted by Subhotosh Khan]:

First we have
If a² + b² = 1 then there is an angle \(\displaystyle \alpha\) so that
sin(\(\displaystyle \alpha\)) = a
and
cos(\(\displaystyle \alpha\)) = b

Next we look at the equation we have
y=3cosx+4sinx
and notice that if we were to divide it by \(\displaystyle \sqrt(3^2+4^2)\) =5 and if we let
a = \(\displaystyle \frac{3}{\sqrt(3^2+4^2)}\) = \(\displaystyle \frac{3}{5}\)
and
b = \(\displaystyle \frac{4}{\sqrt(3^2+4^2)}\) = \(\displaystyle \frac{4}{5}\)
then
a² + b² = 1.
To keep the formula the same, if we divide by something we must multiply by that something, so
y=3cosx+4sinx = 5 (a cos(x) + b sin(x) ) = 5 (sin(\(\displaystyle \alpha\)) cos(x) + cos(\(\displaystyle \alpha\)) sin(x))

The final step is to note that, because of the formula for the sine of the sum of angles,
(sin(\(\displaystyle \alpha\)) cos(x) + cos(\(\displaystyle \alpha\)) sin(x)) = sin(x + \(\displaystyle \alpha\))
and thus
y=3cosx+4sinx = 5 sin(x + \(\displaystyle \alpha\))

To plot the function you need to find out what \(\displaystyle \alpha\) is but you can look that up or use a trig calculator.

 

[mynamesmurph]

OK, I get it now. Thank you very much guys.