1. ## Prove the Identity

$\displaystyle Cos\left(\frac{\pi}{3} + x\right) = \frac{cos(x) - \sqrt{3}\;sin(x)}{2}$

2. $\displaystyle \cos (A + B) = \cos (A)\cos (B) - \sin (A)\sin (B)$
OzzMan, one has to wonder if you are just asking for answers or do you understand any of this.
Which is it?

3. the questions. Im stuck. My teacher is horrible. Same with text book. this is the only place where i learn stuff.

4. Originally Posted by OzzMan
the questions. Im stuck. My teacher is horrible. Same with text book. this is the only place where i learn stuff.
This doesn't answer a lot of "why" question, but does give the information in an organized page.

The teacher may/may not be bad, but I'm betting the text is better than you are giving it credit for. (Probably the teacher as well.) Have you tried talking to your teacher about getting extra help?

-Dan

5. No this teacher is bad. he talks to himself the entire time and lets us out early 1 hour each class. I understand what plato is saying about the identity but i dont get where the 2 in the denominator comes from and as well as where do you go from there.

6. Originally Posted by Plato
$\displaystyle \cos (A + B) = \cos (A)\cos (B) - \sin (A)\sin (B)$
OzzMan, one has to wonder if you are just asking for answers or do you understand any of this.
Which is it?
Originally Posted by OzzMan
$\displaystyle Cos\left(\frac{\pi}{3} + x\right) = \frac{cos(x) - \sqrt{3}\;sin(x)}{2}$
$\displaystyle a = \frac{\pi}{3}$
and
$\displaystyle b = x$

-Dan

7. Originally Posted by OzzMan
$\displaystyle Cos\left(\frac{\pi}{3} + x\right) = \frac{cos(x) - \sqrt{3}\;sin(x)}{2}$
Originally Posted by Plato
$\displaystyle \cos (A + B) = \cos (A)\cos (B) - \sin (A)\sin (B)$
[snip]
Originally Posted by OzzMan
[snip]I understand what plato is saying about the identity but i dont get where you go from there.
You substitute $\displaystyle A = \frac{\pi}{3}$ and $\displaystyle B = x$.

You do know the exact surd values of $\displaystyle \sin \frac{\pi}{3}$ and $\displaystyle \cos \frac{\pi}{3}$ .....?

8. Originally Posted by OzzMan
No this teacher is bad. he talks to himself the entire time and lets us out early 1 hour each class. I understand what plato is saying about the identity but i dont get where you go from there.
If you are certain then you ought to get a tutor of some kind. That is to say, you need to find a warm body in your area to go through this stuff in detail with you.

-Dan

9. I get how $\displaystyle Cos\left(\frac{\pi}{3} + x\right)$ becomes $\displaystyle Cos\left(\frac{\pi}{3}\right)\;Cos(x) - Sin\left(\frac{\pi}{3}\right)\;Sin(x)$

But I dont get how that becomes $\displaystyle \frac{1}{2}\;Cos(x) - \frac{1}{2}\;\sqrt{3}\;Sin(x)$

I know that the Cosine of $\displaystyle \frac{\pi}{3} = .5$ but the sine of $\displaystyle \frac{\pi}{3}$ doesnt equal $\displaystyle .5$

Is this making sense?

10. Originally Posted by OzzMan
I get how $\displaystyle Cos\left(\frac{\pi}{3} + x\right)$ becomes $\displaystyle Cos\left(\frac{\pi}{3}\right)\;Cos(x) - Sin\left(\frac{\pi}{3}\right)\;Sin(x)$

But I dont get how that becomes $\displaystyle \frac{1}{2}\;Cos(x) - \frac{1}{2}\;\sqrt{3}\;Sin(x)$

I know that the Cosine of $\displaystyle \frac{\pi}{3} = .5$ but the sine of $\displaystyle \frac{\pi}{3}$ doesnt equal $\displaystyle .5$

Is this making sense?
$\displaystyle \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$. You must have seen the exact values of sine and cosine of 'special' angles in your previous year of study.

11. Yea. Just was a ton to remember.