# Thread: Trigonomtery Angles: Compound Angle Formulas

1. ## Trigonomtery Angles: Compound Angle Formulas

What would be the formula for
-cos(A-B) = ?
-cos(A+B) = ?
-sin(A-B) = ?
-sin(A+B) = ?

I don't understand what happens with a negative in front of cos and sin, that's my point..

For example:
cos(11pi/12) ends up being -cos(pi/4-pi/6)
cos (17pi/12) ends up being -cos(pi/4-pi/6) also
[correct me if i am wrong]
and then, if i used the formula -cos(A-B) = -cosAcosB+sinAsinB
the formula on the right side of both questions are the same, yet if i leave the same formula on the right for both equations (since the left is the same) then i don't get the right answer.. one of them needs -cosAcosB-sinAsinB for the left and the other one needs -cosAcosB+sinAsinB in order to get the correct answer.

this is what I dont understand, how do you know which one it is?

2. Originally Posted by skeske1234
What would be the formula for
-cos(A-B) = ?
-cos(A+B) = ?
-sin(A-B) = ?
-sin(A+B) = ?

I don't understand what happens with a negative in front of cos and sin, that's my point..
Expand the $\cos{(A - B)}$ (say) as you usually would, then switch all the signs.

3. Originally Posted by Prove It
Expand the $\cos{(A - B)}$ (say) as you usually would, then switch all the signs.
What do you mean..? I don't follow..
Can you try explaining it using my example please and thank you
cos(11pi/12) and cos(17pi/12)

4. Originally Posted by skeske1234
What do you mean..? I don't follow..
Can you try explaining it using my example please and thank you
cos(11pi/12) and cos(17pi/12)
$\cos{\frac{17\pi}{12}} = \cos{\left(\frac{9\pi}{12} + \frac{8\pi}{12}\right)} = \cos{\left(\frac{3\pi}{4} + \frac{2\pi}{3}\right)}$

$\cos{\left(\frac{3\pi}{4} + \frac{2\pi}{3}\right)} = \cos{\left(\frac{3\pi}{4}\right)}\cos{\left(\frac{ 2\pi}{3}\right)} - \sin{\left(\frac{3\pi}{4}\right)}\sin{\left(\frac{ 2\pi}{3}\right)}$

Can you go from here?

For your example re what happens when there's a negative out the front...

$-\cos{\left(\frac{3\pi}{4} + \frac{2\pi}{3}\right)} = -\left[\cos{\left(\frac{3\pi}{4}\right)}\cos{\left(\frac{ 2\pi}{3}\right)} - \sin{\left(\frac{3\pi}{4}\right)}\sin{\left(\frac{ 2\pi}{3}\right)}\right]$

Just use the distributive law...