Mastering Trigonometry: Understanding Sine, Cosine, and Tan Graphs

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Homework Help Overview

The discussion revolves around understanding the sine, cosine, and tangent graphs, particularly in the context of solving problems related to specific angle values and their relationships. Participants are exploring how to derive values from the unit circle and the periodic nature of trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss methods for finding sine values at specific angles without a calculator, referencing the unit circle and the properties of sine functions. Questions arise about identifying multiple solutions for equations involving sine, particularly in the context of the angle transformations.

Discussion Status

There is an active exchange of ideas, with some participants suggesting visual methods such as drawing the unit circle to aid understanding. Others emphasize the importance of memorizing key triangles and their relationships to the unit circle. While some guidance has been offered, there is no explicit consensus on the best approach to solving the problems presented.

Contextual Notes

Participants mention constraints related to non-calculator exams and the need to derive values from fundamental trigonometric principles. There is also a reference to the necessity of understanding the unit circle for solving trigonometric equations.

david18
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I'm having a lot of problems with this topic. I know what the sine, cosine and tan graphs look like.

one question i come across fequently is "given that sin 30°, what is a) sin 150° b) sin 330°. - i can work it out on a calculator but the questions on a non-calc paper. I am presuming it's something about the graph having repetitions above and below the x axis.

another question tat frequently pops up is something like "find two different values of x between 0 and 180 for which sin (2x)° = sin 30°". I can work out one of them as 15° but cannot find the second solution and would like to know how it can be found.
 
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You know how to define cos and sin as the x and y coordinates of points on the unit circle, right? Then it should be easy. Just draw the unit circle and lines through the origin that make 30°, 150° and 330° angles with the x axis. If you know that sin is the y coordinate of the point where the line intersects the unit circle, the answer will be almost obvious once you've drawn the picture.

You solved the first part of the second problem by realizing that two 15° rotations are equal to a 30° rotation. To solve the second part, the first thing you have to do is to find a second point on the unit circle with the y coordinate equal sin 30°. What's the angle between the x-axis and a line from the origin to that point? The answer you seek is half of that.
 
Either brush up on the unit circle or do a bunch of memorizing:
http://img404.imageshack.us/img404/6708/trigbi7.png
 
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If you can draw sin(x) and cos(x) with reasonable accuracy over 1 period, then the answers are obvious from looking at the symmetry of the functions.

Try and draw sin such that it crosses the x-axis at 0,180 and 360 and it has a maximum at 90, minimum at 270 (half-way between 180 and 360).

Draw cos such that it crosses the x-axis at 90 and 270 and has a maximum at 0,360 and a minimum at 180.

I find that to be much easier than memorizing the above formulae.

Fredrik's method might be easier for you. Whatever works best, but it does require some practice.
 
david18 said:
I'm having a lot of problems with this topic. I know what the sine, cosine and tan graphs look like.

one question i come across fequently is "given that sin 30°, what is a) sin 150° b) sin 330°. - i can work it out on a calculator but the questions on a non-calc paper. I am presuming it's something about the graph having repetitions above and below the x axis.

another question tat frequently pops up is something like "find two different values of x between 0 and 180 for which sin (2x)° = sin 30°". I can work out one of them as 15° but cannot find the second solution and would like to know how it can be found.

You should familiarize yourself with the unit circle: http://www.humboldt.edu/~dlj1/PreCalculus/Images/UnitCircle.html

Once you do you'll start to realize things like:
[tex]sin(30) = 1/2[/tex]
[tex]sin(150) = 1/2[/tex]
[tex]sin(330) = -1/2[/tex]

To find sin(2x) = sin(30) over the interval (0,180] becomes simple from above:
sin(2x) = 1/2 and x = 15 and 75 (half of 30 and 150)
 
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Man, these guys are nuts.

All you need to do is to be able to trigonometry with the 30-60-90 and the 45-45-90 triangle. Everything else follows from the unit circle.
 
ZioX said:
Man, these guys are nuts.

All you need to do is to be able to trigonometry with the 30-60-90 and the 45-45-90 triangle. Everything else follows from the unit circle.

What's the difference? If you know trig with those two triangles you've memorized like 80% of the basic unit circle anyways :-p
 

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