The identity of the sum of cos two angles is
[tex]\text{cos(a}+b)=\cos a\cos b-\sin a\sin b[/tex]Since the given expression is
[tex]\cos \frac{3\pi}{7}\cos \frac{9\pi}{28}-\sin \frac{3\pi}{7}\sin \frac{9\pi}{28}[/tex]Compare it with the identity above
[tex]\begin{gathered} a=\frac{3\pi}{7} \\ b=\frac{9\pi}{28} \end{gathered}[/tex]Then the expression can be written as cos (a + b)
[tex]\cos \frac{3\pi}{7}\cos \frac{9\pi}{28}-\sin \frac{3\pi}{7}\sin \frac{9\pi}{28}=\cos (\frac{3\pi}{7}+\frac{9\pi}{28})[/tex]To add the 2 angles equalize their denominators by finding LCM of them
Since LCM of 7 and 28 is 28, then
[tex]\cos (\frac{3\pi}{7}+\frac{9\pi}{28})=\cos (\frac{12\pi}{28}+\frac{9\pi}{28})=\cos (\frac{12\pi+9\pi}{28})=\cos (\frac{21\pi}{28})[/tex]Now we need to find cos (21pi/28)
Since the angle of measures 21pi/28 = 3pi/4 in its simplest form
Since the angle 3pi/4 is greater than pi/2 and pi, then it lies in the second quadrant
The measure of any angle in the second quadrant is between pi/2 and pi, and the value of cos any angle in the second quadrant is negative
[tex]\begin{gathered} \cos (\frac{3\pi}{4})=-\cos \frac{\pi}{4} \\ \cos \frac{\pi}{4}=\frac{\sqrt[]{2}}{2} \\ \cos \frac{3\pi}{4}=-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]The answer is
[tex]-\frac{\sqrt[]{2}}{2}OR-\frac{1}{\sqrt[]{2}}[/tex]
