Evaluation Equations & Inequalities for Trig Functions/Graph  TOPIC_SOLVED

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Evaluation Equations & Inequalities for Trig Functions/Graph

Postby hiroze on Fri Apr 16, 2010 10:56 pm

Hello!

I'm new to the board and am stuck on these types of questions:

Referring to a graph of y = sin x or y = cos x find the exact values of x in the interval [0, 4pi] that satisfy the equation: sin x = 1/2

I know that I am looking for the values of x where the graph of sin x intersects graph of the line y = 1/2, but I don't know how to determine what those values are. I drew the graphs on the same grid, but I don't know what to do from there. Any help is greatly appreciated :D



*Quick note about my background: I haven't had a math class in 10+ years. My last math class was Business Calculus and I've never taken trigonometry. I passed an assessment to get into pre-caclulus (which I am in now) and now that we're on the trig section, I'm behind since it's pretty much review for everyone else and all new material for me.
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Postby stapel_eliz on Fri Apr 16, 2010 11:38 pm

hiroze wrote:Referring to a graph of y = sin x or y = cos x find the exact values of x in the interval [0, 4pi] that satisfy the equation: sin x = 1/2

I'm not sure why they're wanting you to work from the graphs...? This is actually one of reference-angle values they were supposed to have given you to memorize. :wink:

Maybe they want you to find the other instances of sin(x) = 1/2 by using the graph...?
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Re: Evaluation Equations & Inequalities for Trig Functions/G

Postby hiroze on Fri Apr 16, 2010 11:58 pm

Hmmm. I'm not sure what you mean. The answers are pi/6, 5pi/6, 13pi/6, 17pi/6.
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Postby stapel_eliz on Sat Apr 17, 2010 11:38 am

hiroze wrote:Hmmm. I'm not sure what you mean. The answers are pi/6, 5pi/6, 13pi/6, 17pi/6.

What is the basic reference-angle value for "sin(x) = 1/2"?

Looking at the graph of the sine wave over the interval from 0 to 4pi, where is the wave at the same height? :wink:
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Re: Evaluation Equations & Inequalities for Trig Functions/G

Postby hiroze on Sat Apr 17, 2010 7:38 pm

Part of the problem is that since I've never had a trig class, I don't know what I am suppose to memorize and what I am suppose to be able to derive, which is only important since tests are timed, but I've made a point to memorize the reference angles and make sure I can derive them. But anyway, after looking at this for a few days a few things have made sense, but a part of it is still confusing.

I know that the reference angle for sin 1/2 and cos 1/2 is 30 degrees and 60 degrees respectively and tan 1 is 45 degrees. I am able to get the answer pi/6 since 30 degrees equals pi/6. I can also get the answer 13pi/6 because I know that sin has a period of 2pi so pi/6 + 2pi = 13pi/6.

By the same logic the answer 5pi/6 + 2pi = 17pi/6, but I don't know why we can use 5pi/6 as one of the x values on the interval [0, 4pi]. Looking at the picture, I can see that there is an intersection value right before pi, but I don't know why we use 5pi/6. The actual value of that angle is 150 degrees - the reference angle is 30 degrees, but why do we use the reference angle instead of the the 150? Do we use it because it's at the same height (which you mentioned) or is there another reason?

Image

I know I've complicated a simple problem, but it's a years worth of trig inside of a few weeks so every bit of understanding helps :oops:

Thanks for all the help so far though.


Also, and this is not really important, but I think we don't use 7pi/6 or 11pi/6 since those would be negative sine values right?
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Postby stapel_eliz on Sat Apr 17, 2010 9:10 pm

You're exactly right about the height.

You look at the graph and, knowing that sin(x) = 1/2 for x = pi/6 (or for x = 30 degrees; take your pick), you look for the places where the graph looks to be about right for the height. Then you add or subtract 30 degrees (or pi/6) from the closest known angle measure (like pi/2, 360 degrees, etc) to find the value for x. :wink:
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Re:  TOPIC_SOLVED

Postby hiroze on Sat Apr 17, 2010 11:47 pm

stapel_eliz wrote:You're exactly right about the height.

You look at the graph and, knowing that sin(x) = 1/2 for x = pi/6 (or for x = 30 degrees; take your pick), you look for the places where the graph looks to be about right for the height. Then you add or subtract 30 degrees (or pi/6) from the closest known angle measure (like pi/2, 360 degrees, etc) to find the value for x. :wink:


I'm going to go with the height since the adding or subtracting would only work if the graph was to scale. Thank you for your help.
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Postby stapel_eliz on Sun Apr 18, 2010 11:29 am

The subtracting (for the x-values) is sideways (that is, along the x-axis). If x = pi/6 is a solution, then (from the graph of sine) so also is pi - pi/6 = 5pi/6, 2pi + pi/6 = 13pi/6, etc. :wink:
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