Angle Addition Formula for three angles?

1. Aug 24, 2007

PFStudent

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

How do I find,

$$sin\left(\alpha - \beta + \gamma\right) = ???$$

2. Relevant equations

$$sin\left(\alpha\pm\beta\right) = sin\alpha{{.}}cos\beta \pm cos\alpha{{.}}sin\beta$$

and

$$cos\left(\alpha\pm\beta\right) = cos\alpha{{.}}cos\beta \mp sin\alpha{{.}}sin\beta$$

3. The attempt at a solution

$$sin\left(\alpha - \beta + \gamma\right) = ???$$

I know how to do it for four distinct angles,

$$sin\left(\alpha + \beta + \gamma + \psi\right) = ???$$

Where, let

$$\alpha + \beta = \theta$$

$$\gamma + \psi = \phi$$

And then expand, using the earlier identity I mentioned,

$$sin\left(\theta + \phi\right) = sin\theta{{.}}cos\phi + cos\theta{{.}}sin\phi$$

$$sin\left(\theta + \phi\right) = sin(\alpha + \beta){{.}}cos(\gamma + \psi) + cos(\alpha + \beta){{.}}sin(\gamma + \psi)$$

$$sin\left(\theta + \phi\right) = [sin\alpha{{.}}cos\beta + cos\alpha{{.}}sin\beta]{{.}}[cos\gamma{{.}}cos\psi - sin\gamma{{.}}sin\psi] + [cos\alpha{{.}}cos\beta - sin\alpha{{.}}sin\beta]{{.}}[sin\gamma{{.}}cos\psi + cos\gamma{{.}}sin\psi]$$

$$sin\left((\alpha + \beta) + (\gamma + \psi)\right) = [sin\alpha{{.}}cos\beta + cos\alpha{{.}}sin\beta]{{.}}[cos\gamma{{.}}cos\psi - sin\gamma{{.}}sin\psi] + [cos\alpha{{.}}cos\beta - sin\alpha{{.}}sin\beta]{{.}}[sin\gamma{{.}}cos\psi + cos\gamma{{.}}sin\psi]$$

$$sin\left(\alpha + \beta + \gamma + \psi\right) = [sin\alpha{{.}}cos\beta + cos\alpha{{.}}sin\beta]{{.}}[cos\gamma{{.}}cos\psi - sin\gamma{{.}}sin\psi] + [cos\alpha{{.}}cos\beta - sin\alpha{{.}}sin\beta]{{.}}[sin\gamma{{.}}cos\psi + cos\gamma{{.}}sin\psi]$$

However, for three angles, is where I am stumped.

Any help is appreciated.

Thanks,

-PFStudent

Last edited: Aug 24, 2007
2. Aug 24, 2007

Staff: Mentor

If you can do four, why can't you do three?

Let

$$\alpha + \beta = \theta$$

$$\gamma = \gamma$$

3. Aug 24, 2007

PFStudent

Hey,

Thanks for the quicky reply Doc Al, I hesitated to do that because was not sure if the folowing was true,

$$sin\left(\alpha + \beta + \gamma\right) = sin\left((\alpha + \beta) + \gamma\right) = sin\left(\alpha + (\beta + \gamma)\right)$$

The reason I ask is if any quanity (in the parentheses) can be let equal theta and expanded will they all be equal?

That is where I was unsure. That if you took each scenario I mentioned,

$$sin\left((\alpha + \beta) + \gamma\right)$$

$$sin\left(\alpha + (\beta + \gamma)\right)$$

And let the quanity in parentheses equal theta and applied the angle addition formula, would they still all be equal?

Or does it matter which pair of angles you let equal theta (i.e. does the answer change if you pick two different pairs)?

Thanks,

-PFStudent

4. Aug 24, 2007

Dick

Unfortunately, you also did the four angle one wrong. First expand $$sin\left(\theta + \phi\right)$$ and then put the definitions of theta and phi in and keep expanding. Each term should have trig functions of four angles in it. You expanded $$sin\left(\theta \right)+ sin\left(\phi\right)$$.

Last edited: Aug 24, 2007
5. Aug 24, 2007

Staff: Mentor

They better be! (That's the associative property of addition.)

Try it and see!

Thanks for checking, Dick. (I obviously didn't.)

Last edited: Aug 24, 2007
6. Aug 26, 2007

PFStudent

Hey,

Thanks for the help guys, I edited my original post to reflect the correct expansion for angle addition of four angles.

Thanks,

-PFStudent