<s>[INST] Finding $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$ in $\Delta \text{ABC}$ I am attempting the following trigonometry problem. Given that in an acute angled triangle $\Delta \text{ABC}$, the following equalities hold true

$$\cos^2(A)+\cos^2(B)+2\sin(A)\sin(B)\cos(C)=\dfrac{15}{8}\\
\cos^2(B)+\cos^2(C)+2\sin(B)\sin(C)\cos(A)=\dfrac{14}{9}$$
Find the value of $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$.


My Attempt:
Let the unknown quantity be $x$. Then we have, by adding all the terms.
$$2\sum_{cyc}\cos^2(A)-2\sum_{cyc}\sin(A)\sin(B)\cos(A+B)=\dfrac{15}{8}+\dfrac{14}{9}+x$$
Also simplifying the second summation term as follows, we get $$\sin(A)\sin(B)\cos(A+B)=\dfrac{\sin(2A)\sin(2B)}{4}-(1-\cos^2(A))(1-\cos^2(B))$$

I'm not sure how to proceed further. Any hints are appreciated. Even hints to other possible pathways to the solution are welcome. Thanks
 [/INST] Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\cos^2C+\cos^2A+2\sin A\sin B\cos C$$
$$=1+\cos(C+A)\cos(C-A)+...$$
$$=1-\cos B(\cos C\cos A+\sin C\sin A-2\sin C\sin A)$$
$$=1-\cos B\cos(C+A)$$
$$=1+\cos^2B$$
$\implies\cos^2B=\dfrac{15}8-1,\sin B=+\sqrt{1-\cos^2B}=?$
Similarly we can find $\cos^2A$ and hence $\sin A$
Use Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ to find $\cos C$
Can you take it from here?
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