Graphs for trig

[Waliah snowman]

If I wanted to find out the solutions for a sine graph I would find the PV then minus the PV from 180 but if I was finding the solutions for let’s say sin2X it tells me to also minus from 180 again where as 180mis the period of the graph so I’m not understanding why I would still minus it from 180 if now the period of the graph is different

 

[JeffM]

PV means?

”Minus” means “subtract”?

Who is the “it” who tells you things?

 

[Harry_the_cat]

Can you please give an example showing these ideas?

 

[Dr.Peterson]

If I wanted to find out the solutions for a sine graph I would find the PV then minus the PV from 180 but if I was finding the solutions for let’s say sin2X it tells me to also minus from 180 again where as 180mis the period of the graph so I’m not understanding why I would still minus it from 180 if now the period of the graph is different

One reason we need to see an example is that it is not even clear what you are trying to do. There is no such thing as "solutions for a graph"; you appear to be solving a trig equation, but you show no equations at all, just an expression. I assume PV means "principal value of an inverse trig function", but we can talk about this much better with something specific in front of us, such as sin(2x) = 1/2, or whatever. (The period of a graph is not directly related to the work I'd do.)

 

[Waliah snowman]

One reason we need to see an example is that it is not even clear what you are trying to do. There is no such thing as "solutions for a graph"; you appear to be solving a trig equation, but you show no equations at all, just an expression. I assume PV means "principal value of an inverse trig function", but we can talk about this much better with something specific in front of us, such as sin(2x) = 1/2, or whatever. (The period of a graph is not directly related to the work I'd do.)

Let’s say it says sin2X=40 or any number whatever to find out the period you would divide the original period by the coefficient of x isn’t it also if you wanted to find the sin value of a an inverse sine function for a graph the first value would be the actual value of inverse sine of the number so yes the pv as in the principle value and on google and yt vids I watched it says to find the second value on the graph you would do 180 minus the principle value however how can you use that same principle to find the second value when this graph has ½ the period of the orig graph

 

[Waliah snowman]

This would maybe help u understand what kind of question I maybe referring to as this is how u would normally find it out

 

[JeffM]

[MATH]\text {Given: } \alpha \ne 0, \text { and } sin( \alpha x) = p, \text { and } arcsin (p) = y,[/MATH]
where both x and y are measured in degrees. Solve for all values of x.

[MATH]\text {Let } z = \alpha x \implies sin (z) = p \implies arcsin \{ sin(z) \} = arcsin (p) = y \implies[/MATH]
[MATH]z = y + 180k, \text { where } k \text { is any integer.}[/MATH]
[MATH]\therefore \alpha x = y + 180k \implies x = \dfrac{arcsin (p) + 180k}{\alpha}, \text { where } k \text { is any integer}.[/MATH]
Is this what you are asking about?

 

[Dr.Peterson]

Let’s say it says sin2X=40 or any number whatever to find out the period you would divide the original period by the coefficient of x isn’t it also if you wanted to find the sin value of a an inverse sine function for a graph the first value would be the actual value of inverse sine of the number so yes the pv as in the principle value and on google and yt vids I watched it says to find the second value on the graph you would do 180 minus the principle value however how can you use that same principle to find the second value when this graph has ½ the period of the orig graph

First, I hope you realize that sin2X=40 has no solutions, so it is not a good example. I would rather have seen you actually try solving an equation so we can see your own thoughts.

Second, there are multiple ways to solve such an equation; what you see on "Google and YouTube" will be a wide variety of high- and low-quality work from many different sources, so you haven't specified who actually does what you are asking about.

This would maybe help u understand what kind of question I maybe referring to as this is how u would normally find it out

I wouldn't call that "normal" for the kind of problem you are asking about, though maybe it's common somewhere! I don't think I ever use the period of the graph explicitly. My preferred method is to think in terms of the unit-circle definition of the sine. Have you not found any examples of anyone solving an equation with an argument other than x??

But if you want to use this graphical way of thinking, there are two things you could do. Let's suppose you're trying to solve $\sin(2x) = 0.6$ and have found that one solution is $2x = \sin^{-1}(0.6) = 36.87^\circ$, so that $x = 36.87^\circ/2$.

First, you could actually draw the graph of $y = \sin(2x)$, and use the symmetry of THAT graph. It isn't clear to me that you have tried that, which is what I was hoping to see. Since the period is 180 degrees, half a cycle is 90 degrees, and the two solutions would be 36.87 and 90 - 36.87.

Second, you could just think in terms of the graph of $\sin(x)$ as shown in the problem, and see that the two solutions for 2x are $2x = \sin^{-1}(0.6) = 36.87^\circ$ and $2x = 180-\sin^{-1}(0.6) = 180-36.87^\circ$. Then solve both of these equations for x by dividing by 2, and add multiples of the period.

My preferred way is something like JeffM's. We have that 2x is any angle whose sine is 0.6; the latter is either the inverse sine of 0.6 plus any integral multiple of 360, or 180 - the inverse sine, plus any integral multiple of 360. Then just divide each of those by 2. That is, $2x = \sin^{-1}(0.6)+360k$ or $2x = 180-\sin^{-1}(0.6)+360k$; so $x = \sin^{-1}(0.6)/2+180k$ or $x = 90-\sin^{-1}(0.6)/2+180k$.