# Thread: Verifying equivalent trig ratios

1. ## Verifying equivalent trig ratios

Hi,

I hope someone can help.

I'm suppose to prove that sin(2pi/3) = sin(8pi/3) based on the formula that I developed: Sin4x = 4SinxCos-8Sin^3Cosx

This is the textbook solution for the question:

I believe what the solution is trying to say is that we already know sin(2pi/3) since (2pi/3) is a common angle on the unit circle (hence why there was only one step in the textbook solution to know what the ratio is for sin(2pi/3).

Now that I said what I understand, I will now explain what I'm confused about...

1. Can someone please explain to me why 2pi/3 is subbed into the Sin4x equation as x? What is the textbook solution trying to communicate by doing that? I understand how one could sub 2pi/3 into Sin4x as x, but I'm not sure why you would in the first place. Why don't you just sub in 8pi/3 as x?

2. In addition, do you even need to have the equation of Sin4x in order to determine that sin(2pi/3) and sin(8pi/3) are equivalent? For example, I know that ((8 x 180)/3)=480... and 480 - 360 = 120... and 180 - 120 = 60. What I just proved is that the angle 8pi/3 lies within the 2nd quadrant and has a related acute angle of 60 degrees... which is the same as 2pi/3! Hopefully what I just said didn't sound too convoluted...

Sincerely,
Olivia

2. ## Re: Verifying equivalent trig ratios

1) $4\cdot \dfrac{2\pi}{3} = \dfrac{8\pi}{3}$. This way you get a relationship between the two terms you want to determine equivalence.
2) there are many ways to prove the same thing. It is important in mathematics to know and understand as many techniques as possible to maximize your problem-solving arsenal.

3. ## Re: Verifying equivalent trig ratios

Are you saying that the benefit in using Sin4x is that we can immediately see the relationship between 2pi/3 and 8pi/3 once we plug in 2pi/3 as x into Sin4x? The relationship is that 8pi/3 is 4 times the size of 2pi/3.

It almost seems that there is really no point to further solve Sin4x once you already notice this relationship... isn't that true?

4. ## Re: Verifying equivalent trig ratios

"We can immediately see the relationship between 2pi/3 and 8pi/3"? The crucial relationship, I would think, is that 8pi/3= (6+ 2)pi/3= 6pi/3+ 2pi/3= 2pi+ 2pi/3! Sine and cosine are periodic with period 2pi so it immediately follows that sin(8pi/3)= sin(2pi/3) and cos(8pi/3)= cos(2pi/3).

5. ## Re: Verifying equivalent trig ratios

Suppose we need to calculate $\sin(x)$ and $\sin(4x)$ for non-standard values of $x$. For example, we are trying to run a simulation where we need a high degree of precision. The algorithm takes hours to calculate $\sin(\sqrt{\pi})$ to 500,000 digits. We then need $\sin(4\sqrt{\pi})$ to the same precision. We can either use the same algorithm to calculate 500,000 digits of this calculation, or we can use known relationships to make it faster to calculate.

6. ## Re: Verifying equivalent trig ratios

Originally Posted by SlipEternal
Suppose we need to calculate $\sin(x)$ and $\sin(4x)$ for non-standard values of $x$. For example, we are trying to run a simulation where we need a high degree of precision. The algorithm takes hours to calculate $\sin(\sqrt{\pi})$ to 500,000 digits. We then need $\sin(4\sqrt{\pi})$ to the same precision. We can either use the same algorithm to calculate 500,000 digits of this calculation, or we can use known relationships to make it faster to calculate.
You're saying that because we are aware of the relationship between Sinx and Sin4x, that we can choose to use Sin4x in certain situation in order to make computations more efficient?

7. ## Re: Verifying equivalent trig ratios

See my example in the other thread where you are discussing the same thing.

8. ## Re: Verifying equivalent trig ratios

Originally Posted by HallsofIvy
"We can immediately see the relationship between 2pi/3 and 8pi/3"? The crucial relationship, I would think, is that 8pi/3= (6+ 2)pi/3= 6pi/3+ 2pi/3= 2pi+ 2pi/3! Sine and cosine are periodic with period 2pi so it immediately follows that sin(8pi/3)= sin(2pi/3) and cos(8pi/3)= cos(2pi/3).
It seems like you're saying that if we start at 2pi/3 and take 4 increments of 2pi/3, that would result in a full cycle of Sinx or Cosx, right? Hence why 8pi/3 = 2pi + 2pi/3. In other words, Sin4(2pix/3) = Sin(8pi/3).

9. ## Re: Verifying equivalent trig ratios

I'm not sure what you are asking anymore. I think you may be missing the point. These formulas are a way to calculate sin (4x) when all you know is sin x or cos x. They give you an example where you already know both sin x and sin (4x) to see how it works. See my example in the other thread for a problem where you would not have an easy way to calculate sin (4x) because we have not even told you specifically what x is.

10. ## Re: Verifying equivalent trig ratios

Originally Posted by otownsend
It seems like you're saying that if we start at 2pi/3 and take 4 increments of 2pi/3, that would result in a full cycle of Sinx or Cosx, right? Hence why 8pi/3 = 2pi + 2pi/3. In other words, Sin4(2pix/3) = Sin(8pi/3).
It's rather trivial that 4(2)= 8. No, that was NOT my point. My point was that sin(8pi/3)= sin(2pi/3) follows immediately from the fact that sine is periodic with period 2pi.