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.
Sum and Differences Identity
- Thread starter mchase
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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.
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- Nov 24, 2012
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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:
\(\displaystyle \sin\left(\dfrac{17\pi}{12} \right)=-\dfrac{\sqrt{2+\sqrt{3}}}{2}=-\dfrac{\sqrt{2}(1+\sqrt{3})}{4}\)
\(\displaystyle \sin\left(\dfrac{17\pi}{12} \right)\)
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:
\(\displaystyle \sin\left(\dfrac{17\pi}{12} \right)=-\dfrac{\sqrt{2+\sqrt{3}}}{2}=-\dfrac{\sqrt{2}(1+\sqrt{3})}{4}\)