The identity of the sum of cos two angles is
![\text{cos(a}+b)=\cos a\cos b-\sin a\sin b](https://img.qammunity.org/2023/formulas/mathematics/college/k3i56c20beeetz9al72ie37pf34mkd5e63.png)
Since the given expression is
![\cos (3\pi)/(7)\cos (9\pi)/(28)-\sin (3\pi)/(7)\sin (9\pi)/(28)](https://img.qammunity.org/2023/formulas/mathematics/college/4yu2egbn66ptx94w9iri9q0gu23zyao0me.png)
Compare it with the identity above
![\begin{gathered} a=(3\pi)/(7) \\ b=(9\pi)/(28) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gz8ni11euvzbti3qr8s5s152ty78mxu9co.png)
Then the expression can be written as cos (a + b)
![\cos (3\pi)/(7)\cos (9\pi)/(28)-\sin (3\pi)/(7)\sin (9\pi)/(28)=\cos ((3\pi)/(7)+(9\pi)/(28))](https://img.qammunity.org/2023/formulas/mathematics/college/ccss28xe2hwspuai3ne04ar7d6estkfut4.png)
To add the 2 angles equalize their denominators by finding LCM of them
Since LCM of 7 and 28 is 28, then
![\cos ((3\pi)/(7)+(9\pi)/(28))=\cos ((12\pi)/(28)+(9\pi)/(28))=\cos ((12\pi+9\pi)/(28))=\cos ((21\pi)/(28))](https://img.qammunity.org/2023/formulas/mathematics/college/fa1gsc7agxmxvy88sqpe27oq02uucfim5r.png)
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
![\begin{gathered} \cos ((3\pi)/(4))=-\cos (\pi)/(4) \\ \cos (\pi)/(4)=\frac{\sqrt[]{2}}{2} \\ \cos (3\pi)/(4)=-\frac{\sqrt[]{2}}{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hwdd4tfmhapbhr3wooy8r3cajvchglwdt2.png)
The answer is
![-\frac{\sqrt[]{2}}{2}OR-\frac{1}{\sqrt[]{2}}](https://img.qammunity.org/2023/formulas/mathematics/college/rtysnx9mml4as7fex0twopqpaelv0a498r.png)