Finding $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$ in $\Delta \text{ABC}$

Question

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 $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$.

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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 $$\sin(A)\sin(B)\cos(A+B)=\dfrac{\sin(2A)\sin(2B)}{4}-(1-\cos^2(A))(1-\cos^2(B))$$

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I'm not sure how to proceed further. Any hints are appreciated. Even hints to other possible pathways to the solution are welcome. Thanks

Answer 1

Hint: Note that $$2\sin A\sin B\cos C-2\cos A\cos B\cos C=2\cos(180^\circ-A-B)\cdot\cos C=2\cos^2 C$$

So, the first equation becomes $$\cos^2 A+\cos^2 B+\cos^2 C+2\cos A\cos B\cos C=\frac{15}8$$This is symmetric, which makes it quite useful.

Now, we subtract the second equation from this to get $$\cos^2 A+2\cos A\cos B\cos C-2\sin B\sin C\cos A=\frac{15}8-\frac{14}9$$$$=\cos^2 A+2\cos(B+C)\cos A$$You can take it from here!

Answer 2

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?