# Proving trigonometric identities using sum and difference formulas

1. Jan 4, 2009

### meeklobraca

1. The problem statement, all variables and given/known data

Prove that sin (a+b) does not equal sin a + sin b. Let a = pi/3 and b = pi/6

2. Relevant equations

3. The attempt at a solution

Where im lost is how to figure out what pi/3 or pi/6 is. Like for example, how can I know sin pi/3 is sqrt3 over 2?

Is this a case where I have to memorize what they equal to figure it out, or is there a way I can calculate that. Also any kind of reference material to this would be helpful as I need to brush up on this subject.

Thank You!

2. Jan 4, 2009

### jgens

Are you familiar with the solutions for a few general angles (30, 45, 60, etc.) when the argument for sine and cosine is in degrees?

Ex: pi/3 radians = 60 degrees and pi/6 degrees = 30 degrees. Can you take it from there?

3. Jan 4, 2009

### meeklobraca

Im vagely familar with it, but thats where I need to brush up on the subject.

Like I know pi is 180 degrees, hence why that divided by 3 = 60 degrees, but I dont know why sin pi/3 = sqrt3 over 2 or why sin pi/2 = 1 but cos pi/2 = 0

4. Jan 4, 2009

### ko12sd

Our teacher had us memorize the unit circle in both degrees and radians. I think this is pretty common, but anyway there's one on Wikipedia that gives it in both measures. Here's a link: http://en.wikipedia.org/wiki/Unit_circle. (x,y) = (cos(t), sin(t)) It's huge and about halfway down the page, can't miss it ;).

5. Jan 4, 2009

### jgens

Well, if you're familiar with the unit circle definition of the trigonometric functions, the sine and cosine of them aren't altogether that difficult to calculate.

Definition: We define the trigonometric functions as following: sin(theta) = y/r and cos(theta) = x/r where x and y refer to specific points on the unit circle and r is the radius. All other trig. functions can be defined as variants of these.

We define the unit circle as following: x^2 + y^2 = 1.

Hence, suppose we wish to find the sin(pi/2). Examining the unit circle - in standard position - we find an arc of pi/2 occurs at the point (0,1); therefore, by our definition of trig. functions sin(pi/2) = 1. Note that this really only works well for the arcs pi/2, pi/4, and 0. Does that make sense?

Edit: Now, here's a proof for the values of pi/3 and pi/6. Consider a 30-60-90 triangle. We may claim that the area of of 30-60-90 triangle is half that of an equilateral triangle. If an equilateral triangle is of side length 1 (as is the case in the unit circle) it is then immediately obivous that the side adjacent the 60 degree angle measure is 1/2. Therefore, cos(60) = cos(pi/3) = 1/2. Similar logic will yield the values for sin(pi/3) and all the other related values.

Sorry if the above doesn't make sense. Try drawing it out, it should help.

Last edited: Jan 4, 2009
6. Jan 4, 2009

### symbolipoint

The best approach to the question should not be to substitute numeric values, but to attempt a comparison between the known identity and the proposed identity. Try to carry steps to show the two expressions are equal. A show of equality should not be possible.

Your hope of using actual values is not bad; it is just less general.

7. Jan 4, 2009

### Dick

I really strongly disagree with that. Trying to show something is false because you can't prove it is a hopeless enterprise. A counterexample is a perfectly general proof that an identity is false.

8. Jan 5, 2009

### HallsofIvy

Staff Emeritus
Yes, 180/3= 60 so an equilateral triangle has angles of 60 degrees. If you draw a perpendicular from one vertex to the opposite side, you divide it into two right triangles with angles of 90, 60 and 30 degrees. If you take one side of the equilateral triangle to be 1, the right triangles have hypotenuse of length 1 and one leg, opposite the 30 degree angle, of 1/2. That tells you immediately that sin(30)= opposite/hypotenuse= 1/2 and, of course, cos(30)= 1/2. Now you can use the Pythagorean theorem to find the length of the other leg (the altitude of the equilateral triangle) and so find sin(60) and cos(30).