Finding the Intersection of a Sinusoidal Function and a Line

AI Thread Summary
To find the intersection points of the sinusoidal function y=sin(x) and the line y=x/7, graphing both functions is recommended for an approximate solution. The obvious intersection occurs at x=0, but additional intersections require numerical methods, as algebraic techniques are ineffective. Newton's method is suggested for finding these intersections, although it may not always converge reliably. The discussion also notes that similar methods can be applied to other functions like y=tan(x) and y=cos(x). Understanding these concepts will be beneficial for future calculus studies.
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Homework Statement


Hi! I'm trying to find the points of intersection of a sinusoidal function and a line. The line is y=x/7. The function is y=sinx. Can someone tell me how to determine the number of intersections and exact intersections. I would also like to know if the same method can be applied to y=tanx and y=cosx. Thanks! Also, can you keep the math at a high school level; pre-calculus.

Homework Equations

The Attempt at a Solution

 
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SimmonSays said:

Homework Statement


Hi! I'm trying to find the points of intersection of a sinusoidal function and a line. The line is y=x/7. The function is y=sinx. Can someone tell me how to determine the number of intersections and exact intersections. I would also like to know if the same method can be applied to y=tanx and y=cosx. Thanks! Also, can you keep the math at a high school level; pre-calculus.

Homework Equations

The Attempt at a Solution

Working with trig functions means that most algebraic techniques for finding intersections can't be used.

If you want to find out approximately where y = x/7 and y = sin (x) intersect, probably the quickest way is to make a graph of these two functions.

There should be one obvious value of x which satisfies these two equations. Locating the others is a little more difficult and probably will require some trial and error calculations to find the values of x.
 
y= x/7 and y= sin(x) give x/7= sin(x). An obvious solution to that is x= 0, As for any other solution (and there are other solutions, the slope of x/7 is 1/7 while the slope of sin(x), at x= 0 is 1 so the line goes under the sine curve- but the sine curve turns back down again), as SteamKing says, there is no "algebra way" to solve that, you will need to solve it numerically. One method that tends to converge fairly quickly is "Newton's" method: to solve f(x)= 0, choose some starting value, x_0, construct the next value, x_1= x_0- \frac{f(x_0)}{f'(x_0)}. Then construct x_2= x_1- \frac{f(x_1)}{f'(x_1)} and continue like that until you have sufficient accuracy.

Here, the problem is to solve f(x)= x/7- sin(x)= 0 so f'(x)= 1/7- cos(x) and the formula becomes x_{n+1}= x_n- \frac{x_n/7- sin(x_n)}{1/7- cos(x_n)}= x_n- \frac{x_n- 7sin(x_n)}{1- 7cos(x_n)}.
 
Thank you SteamKing and HallsofIvy for the replies. It actually helped a lot. I'll graph the two functions from now on to find the intersection. I was generally interested in the mathematics if there were an algebraic way to solve. Oh, I didn't know about Newton's method previously; thank you for telling me about. I do not understand it that well at the moment; maybe I will use it in calculus or high-level mathematics. Thank you though.
 
SimmonSays said:
Thank you SteamKing and HallsofIvy for the replies. It actually helped a lot. I'll graph the two functions from now on to find the intersection. I was generally interested in the mathematics if there were an algebraic way to solve. Oh, I didn't know about Newton's method previously; thank you for telling me about. I do not understand it that well at the moment; maybe I will use it in calculus or high-level mathematics. Thank you though.
A word of caution on Newton's method: it does not always converge. If you tried to use it to find where x1/3 becomes zero, starting at some nonzero value for x, you would find each subsequent x value is -2 times the previous one, taking you further and further from the answer. There are ways around this.
 
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