Lesson Using 30,60,90 Triangles and ASTC to solve trig equations

Source code of 'Using 30,60,90 Triangles and ASTC to solve trig equations'

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Consider a 30,60,90 triangle.

We know from converting radians and degrees, that these triangles are equivalent.

Here is it in radian form:

This second triangle is how the {{{pi/6}}} {{{pi/3}}} values are formed on the unit circle.
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Now consider a 45,45,90 triangle.

Its radian counterpart is:
<img src="http://i57.photobucket.com/albums/g215/swincher4391/th_45-45.png?t=1304540166"/>

This is how the {{{pi/4}}} values are formed.
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Now just remember SOH-CAH-TOA,  sin = opp/hypo   cos = adj/hypo  and tangent = opp/adj,
and you should be able to find any of these values using the triangles.
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Remember (x,y) = {{{cos(theta)}}},{{{sin(theta)}}}

Consider the quadrangles from [{{{0}}}, {{{2pi}}})

Then by looking at the points, we can figure out easily what the values are at those angles.

So we have the fundamental angles and their values... how do we extend it to {{{2pi}}}?

We need to know two things. The reference angle and the starting angle.

Step 1) Find what quadrant the angle you are given is.  
Step 2) Starting with your angle find the nearest angle that lies on the x-axis. 
These are the multiples of pi (0,pi,2pi,3pi) etc...
Step 3) Going counterclockwise, if the angle is behind you, take your starting angle and subtract your x-axis angle.  If it is in front of you, take the x-axis angle and subtract from it the starting angle. (Note 0 = 2PI) This new angle is your reference angle. It should between 0 and {{{pi/2}}}.
Step 4) Use this as your new value to find the sin, cos, tan.. etc of.
Step 5) Determine whether your value should be positive or negative. Use the chart below to determine.
*** Note : A- All positive, S- sin positive, T- tangent (which implies sin and cos are negative), C- cosine ***

That's it!

Let's go through an example.

{{{sin((5pi)/3) = x}}}

Step 1)What quadrant is {{{(5pi)/3}}} in? Is it bigger than {{{(3pi)/2}}}? Yes, just slightly. So it is in the 4th quadrant.

Step 2) The closest angle would be in front of me at {{{2pi}}}.

Step 3) So I take {{{2pi - 5pi/3}}} and I get{{{highlight(pi/3)}}}.

Step 4) Look at the triangle to see that {{{sin(pi/3)}}} = {{{sqrt(3)/2}}}.

Step 5) Reference ASTC. In Q4 we have a C. This means cosine is positive, sine is negative. We have sin, so our answer is negative.

The answer then is {{{-sqrt(3) /2 }}}.

Let's ask the opposite question now.

{{{sin(x) = -sqrt(3)/2}}}.

Go in the exact opposite order.

One might be inclined to say that the answer is {{{(5pi)/3}}}. True, that is one answer. Let's make sure there aren't others.

Look at our triangle.  At which angle is {{{sin(x) = sqrt(3)/2}}}?

{{{pi/3}}}

So we ask this question, in what quadrants is sin negative? We look and we see that in Q3 and Q4 sin is negative.

Locate the x-axis angle in Q3 first: it is {{{pi}}}.

So what is {{{(pi)/3 + pi}}}? (We add cause we want to move forward to stay in the 3rd quadrant) {{{highlight((4pi)/3)}}}.

Locate the x-axis angle in Q4: it is {{{2pi}}}.

So what is {{{2pi - pi/3}}}? (We subtracted because we want to go backwards from 2pi to stay in the 4th quadrant) It is {{{highlight((5pi)/3)}}}.

So our answers are {{{(4pi)/3}}} {{{(5pi)/3}}}.

I hope this helped!