We are given the following expression:
[tex]\sin (\frac{3\pi}{4}+\frac{2\pi}{3})[/tex]
the identity for the sum of angles for sines is the following:
[tex]\sin (\alpha+\beta)=\sin \alpha\cos \beta+\cos \alpha\sin \beta[/tex]
In this case, we have:
[tex]\begin{gathered} \alpha=\frac{3\pi}{4} \\ \\ \beta=\frac{2\pi}{3} \end{gathered}[/tex]
Substituting in the identity we get:
[tex]\sin (\frac{3\pi}{4}+\frac{2\pi}{3})=\sin (\frac{3\pi}{4})\cos (\frac{2\pi}{3})+\cos (\frac{3\pi}{4})\sin (\frac{2\pi}{3})[/tex]
we have the following value:
[tex]\sin (\frac{3\pi}{4})=\frac{1}{\sqrt[]{2}}[/tex][tex]\cos (\frac{3\pi}{4})=-\frac{1}{\sqrt[]{2}}[/tex][tex]\sin (\frac{2\pi}{3})=\frac{\sqrt[]{3}}{2}[/tex][tex]\cos (\frac{2\pi}{3})=-\frac{1}{2}[/tex]
Now, we substitute the values in the indentity:
[tex]\sin (\frac{3\pi}{4})\cos (\frac{2\pi}{3})+\cos (\frac{3\pi}{4})\sin (\frac{2\pi}{3})=(\frac{1}{\sqrt[]{2}})(-\frac{1}{2})+(-\frac{1}{\sqrt[]{2}})(\frac{\sqrt[]{3}}{2})[/tex]
Simplifying we get:
[tex](\frac{1}{\sqrt[]{2}})(-\frac{1}{2})+(-\frac{1}{\sqrt[]{2}})(\frac{\sqrt[]{3}}{2})=-\frac{1}{2\sqrt[]{2}}-\frac{\sqrt[]{3}}{2\sqrt[]{2}}[/tex]
Solving the operations:
[tex]-\frac{1}{2\sqrt[]{2}}-\frac{\sqrt[]{3}}{2\sqrt[]{2}}=-0.97[/tex]
Therefore, the value of the sine is -0.97
