Sum and Differences Identity

[mchase]

I'm having issues understanding the Sum and Differences Identity. I understand the concepts, and I know the formula, I'm just not understanding what figures I have to sub in for alpha and beta. For instance, I'm not sure how to answer this question, 17pi/12. I know what to do once I have the two pi figures (like pi/4, or 11pi/6, etc of the unit circle), but I have no idea how to find the right units. I would love some help with this.

 

[wjm11]

I'm having issues understanding the Sum and Differences Identity. I understand the concepts, and I know the formula, I'm just not understanding what figures I have to sub in for alpha and beta. For instance, I'm not sure how to answer this question, 17pi/12. I know what to do once I have the two pi figures (like pi/4, or 11pi/6, etc of the unit circle), but I have no idea how to find the right units. I would love some help with this.

17pi/12 is NOT a question. It is an angle expressed in radians. Please post the complete problem statement.

Also, it would be helpful if you read the "Read before posting" notice.

 

[MarkFL]

Suppose you are told to find the value of:

$\displaystyle \sin\left(\dfrac{17\pi}{12} \right)$

You may observe that:

$\displaystyle \dfrac{17\pi}{12}=\dfrac{18\pi-\pi}{12}=\dfrac{3\pi}{2}-\dfrac{\pi}{12}$ so we have:

$\displaystyle \sin\left(\dfrac{17\pi}{12} \right)=\sin\left(\dfrac{3\pi}{2}-\dfrac{\pi}{12} \right)$

Now, using the angle-difference identity for sine, we may write:

$\displaystyle \sin\left(\dfrac{17\pi}{12} \right)=\sin\left(\dfrac{3\pi}{2} \right)\cos\left(\dfrac{\pi}{12} \right)-\cos\left(\dfrac{3\pi}{2} \right)\sin\left(\dfrac{\pi}{12} \right)$

Since $\displaystyle \sin\left(\dfrac{3\pi}{2} \right)=-1$ and $\displaystyle \cos\left(\dfrac{3\pi}{2} \right)=0$ we now have:

$\displaystyle \sin\left(\dfrac{17\pi}{12} \right)=-\cos\left(\dfrac{\pi}{12} \right)$

Using the half-angle identity for cosine, we have:

$\displaystyle \cos\left(\dfrac{\pi}{12} \right)=\sqrt{\dfrac{1+\cos\left(\dfrac{\pi}{6} \right)}{2}}=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$

and thus we have:

$\displaystyle \sin\left(\dfrac{17\pi}{12} \right)=-\dfrac{\sqrt{2+\sqrt{3}}}{2}=-\dfrac{\sqrt{2}(1+\sqrt{3})}{4}$