Assume that we are working with angles in degrees.
Hint:
$$\cos(195) = \cos(180 + 15)$$
Another hint:
Now split $\cos(180 + 15)$ using $\cos(A+B) = \cos(A)\cos(B)-\sin(A)\sin(B)$.
And another:
Substitute $\sin(180) = 0$ and $\cos(180) = -1$.
Nearly done:
Notice that $\cos(15) = \cos(30/2)$ and remember that $\cos(2A) = 2\cos^{2}(A) - 1$.
Now draw the 30/60/90 triangle and evaluate.
Observe that as $\displaystyle0< A<45^\circ, 0<\tan A<\tan45^\circ=1$
So,we have $\displaystyle\tan A+\frac1{\tan A}=\frac{841}{420}$
$\displaystyle\iff420\tan^2A-841\tan A+420=0$
So, the discriminant will be $\displaystyle841^2-4\cdot420\cdot420=841^2-840^2=841+840=41^2$
Solve the Quadratic Equation for $\tan A$ to get $\displaystyle\tan A=\frac{841\pm41}{420}$
Clearly, $\displaystyle\tan A=\frac{841+41}{2\cdot420}>1$
So, we need $\displaystyle\tan A=\frac{841-41}{2\cdot420}=\cdots$ as it lies in our required range
Also, $\displaystyle\sin A>0, \cos A>0$
So, $\displaystyle\cos A=+\frac1{\sqrt{\sec^2A}}=+\frac1{\sqrt{1+\tan^2A}}$ and $\displaystyle\tan A=\frac{\sin A}{\cos A}\iff\sin A=\tan A\cos A $
Best Answer
Note that, $$ \cos(\alpha\pm \beta)=\cos \alpha\cos \beta\mp\sin \alpha\sin \beta $$ and $$ \cos(180^\circ-\alpha)=-\cos\alpha. $$ Hence $$ \begin{align} \cos52^\circ+\cos68^\circ+\cos172^\circ&=\color{blue}{\cos(60^\circ-8^\circ)+\cos(60^\circ+8^\circ)}+\color{red}{\cos(180^\circ-8^\circ)}\\ &=\color{blue}{2\cos60^\circ\cos8^\circ}+(\color{red}{-\cos8^\circ})\\ &=\color{blue}{2\cdot\frac12\cdot\cos8^\circ}-\color{red}{\cos8^\circ}\\ &=0. \end{align} $$