\[cos\text{ }55{}^\circ \text{ }+\text{ }cos\text{ }65{}^\circ \text{ }+\text{ }cos\text{ }175{}^\circ \text{ }=\text{ }0\]
Take LHS:
\[cos\text{ }55{}^\circ \text{ }+\text{ }cos\text{ }65{}^\circ \text{ }+\text{ }cos\text{ }175{}^\circ \]
we know that:
\[cos\text{ }A\text{ }+\text{ }cos\text{ }B\text{ }=\text{ }2\text{ }cos\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2\]
\[cos\text{ }55{}^\circ \text{ }+\text{ }cos\text{ }65{}^\circ \text{ }+\text{ }cos\text{ }175{}^\circ \text{ }=\text{ }2\text{ }cos\text{ }({{55}^{o}}~+\text{ }{{65}^{o}})/2\text{ }cos\text{ }({{55}^{o}}~\text{ }{{65}^{o}})\text{ }+\text{ }cos\text{ }({{180}^{o}}~\text{ }{{5}^{o}})\]
\[=\text{ }2\text{ }cos\text{ }{{120}^{o}}/2\text{ }cos\text{ }(-{{10}^{o}})/2\text{ }\text{ }cos\text{ }{{5}^{o}}~\left( since,\text{ }\{cos\text{ }\left( 180{}^\circ \text{ }\text{ }A \right)\text{ }=\text{ }\text{ }cos\text{ }A\} \right)\]
\[=\text{ }2\text{ }cos\text{ }60{}^\circ \text{ }cos\text{ }\left( -5{}^\circ \right)\text{ }\text{ }cos\text{ }5{}^\circ \text{ }\left( since,\text{ }\{cos\text{ }\left( -A \right)\text{ }=\text{ }cos\text{ }A\} \right)\]
\[=\text{ }2\text{ }\times \text{ }1/2\text{ }\times \text{ }cos\text{ }{{5}^{o}}~\text{ }cos\text{ }{{5}^{o}}\]
\[=\text{ }0\]
= RHS
Hence Proved.