91Ó°ÊÓ

By looking at the given angles, it can be observed that we can express \(\frac{13 \pi}{12}\) as the sum of \(\frac{7 \pi}{12}\) and \(\frac{6 \pi}{12}\). Thus, write: $$ \frac{13 \pi}{12}=\frac{7 \pi}{12}+\frac{6 \pi}{12} $$ Now, we can observe that the angle \(\frac{7 \pi}{12}\) can be expressed as the difference of \(\frac{\pi}{12}\) and \(\frac{\pi}{8}\), and the angle \(\frac{6 \pi}{12}\) is equal to \(\frac{1}{2} \pi\). So the expression will become: $$ \frac{13 \pi}{12}=\left(\frac{\pi}{12}-\frac{\pi}{8}\right) + \frac{1}{2} \pi $$

Now that we have expressed the angle \(\frac{13 \pi}{12}\) as the combination of the given angles, we can apply trigonometric identities. Using sum and difference identities, the given expression can be rewritten as: $$ \sin \frac{13 \pi}{12} = \sin \left[\left(\frac{\pi}{12}-\frac{\pi}{8}\right) + \frac{1}{2} \pi\right] = \sin \left(\frac{\pi}{12}-\frac{\pi}{8}\right) \cos \left(\frac{1}{2} \pi\right) + \cos \left(\frac{\pi}{12}-\frac{\pi}{8}\right) \sin \left(\frac{1}{2} \pi\right) $$ Now, we apply again the difference formula for sine and cosine: $$ \sin \left(\frac{\pi}{12}-\frac{\pi}{8}\right)=\sin\frac{\pi}{12}\cos\frac{\pi}{8}-\cos\frac{\pi}{12}\sin\frac{\pi}{8} $$ $$ \cos \left(\frac{\pi}{12}-\frac{\pi}{8}\right)=\cos\frac{\pi}{12}\cos\frac{\pi}{8}+\sin\frac{\pi}{12}\sin\frac{\pi}{8} $$ Since \(\cos\left(\frac{1}{2}\pi\right)=0\), and \(\sin\left(\frac{1}{2}\pi\right)=1\), the expression becomes: $$ \sin\frac{13\pi}{12}=\cos \left(\frac{\pi}{12}-\frac{\pi}{8}\right) = \cos\frac{\pi}{12}\cos\frac{\pi}{8}+\sin\frac{\pi}{12}\sin\frac{\pi}{8} $$

Now, we will plug in the given values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\): $$ \sin\frac{13\pi}{12}=\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)\cos\frac{\pi}{8}+\left(\frac{\sqrt{2-\sqrt{2}}}{2}\right)\sin\frac{\pi}{8} $$ To find the value of \(\cos\frac{\pi}{8}\) and \(\sin\frac{\pi}{8}\), we can use the Pythagorean identity \(\cos^2\alpha+\sin^2\alpha=1\). Therefore: $$ \cos\frac{\pi}{8}=\sqrt{1-\sin^2\frac{\pi}{8}}=\sqrt{1-\left(\frac{\sqrt{2-\sqrt{2}}}{2}\right)^2}=\frac{\sqrt{2+\sqrt{2}}}{2} $$ Now plug the value of \(\cos\frac{\pi}{8}\) into the expression: $$ \sin\frac{13\pi}{12}=\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)\left(\frac{\sqrt{2+\sqrt{2}}}{2}\right)+\left(\frac{\sqrt{2-\sqrt{2}}}{2}\right)\left(\frac{\sqrt{2-\sqrt{2}}}{2}\right) $$

Finally, we will simplify the expression for \(\sin\frac{13\pi}{12}\): $$ \sin\frac{13\pi}{12}=\frac{1}{4}\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{2}}\right)+\frac{1}{4}\left(\sqrt{2-\sqrt{2}}\right)^2 $$ $$ \sin\frac{13\pi}{12}=\frac{1}{4}\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{2}}\right)+\frac{1}{4}\left(1\right) $$ $$ \sin\frac{13\pi}{12}=\frac{1}{4}\left[\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{2}}\right)+1\right] $$ The exact value for the sine of \(\frac{13\pi}{12}\) can be expressed as: $$ \sin\frac{13\pi}{12}=\frac{1}{4}\left[\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{2}}\right)+1\right] $$