N 23
we have
[tex]sin(\frac{3pi}{4}+\frac{5pi}{6})[/tex]Remember that
[tex]sin\left(a+b\right)=sin\left(a\right)cos\left(b\right)+cos\left(a\right)\lparen sin\left(b\right)[/tex]substitute given values
[tex]s\imaginaryI n(\frac{3pi}{4}+\frac{5pi}{6})=s\imaginaryI n(\frac{3p\imaginaryI}{4})cos(\frac{5p\imaginaryI}{6})+cos(\frac{3p\imaginaryI}{4}\operatorname{\lparen}s\imaginaryI n(\frac{5p\imaginaryI}{6})[/tex]Remember that
3pi/4=135 degrees ------> reference angle is 45 degrees
135 degrees -----> II quadrant ----> sine is positive and cosine is negative
so
sin(3pi/4)=sin(45)=√2/2
cos(3pi/4)=-cos(45)=-√2/2
5pi/6=150 degrees ----> reference angle is 30 degrees
150 degrees -----> II quadrant -----> sine is positive and cosine is negative
sin(5pi/6)=sin(30)=1/2
cos(5pi/6)=-cos(30)=-√3/2
Substitute the given values in the formula
[tex]s\imaginaryI n(\frac{3p\imaginaryI}{4}+\frac{5p\imaginaryI}{6})=\frac{\sqrt{2}}{2}*(-\frac{\sqrt{3}}{2})+(-\frac{\sqrt{2}}{2})*(\frac{1}{2})[/tex][tex]s\imaginaryI n(\frac{3p\imaginaryI}{4}+\frac{5p\imaginaryI}{6})=-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4})[/tex]The answer is
[tex]s\imaginaryI n(\frac{3p\imaginaryI}{4}+\frac{5p\imaginaryI}{6})=\frac{-\sqrt{6}-\sqrt{2}}{4}[/tex]