Sum of cosines of complementary/suplementary angles Why are $(\cos(2^{\circ})+\cos(178^{\circ})), (\cos(4^{\circ})+\cos(176^{\circ})),.., (\cos(44^{\circ})+\cos(46^{\circ}))$ all equal zero?
Could you prove it by some identity?
 A: 
so 
  $$cos (2) +cos(178) =cos(2)+cos(180-2)=0\\ cos (4) +cos(176) =cos(4)+cos(180-4)=0\\...\\$$
A: Three relevant identities:
$\cos(90^{\circ} - x) = \sin(x)$,
$\sin(90^{\circ} - x) = \cos(x)$,
$\sin(-x) = - \sin(x)$.
Therefore,
\begin{align*}
\cos(180^{\circ} - x) &= \cos(90^{\circ} - (x - 90^{\circ}))\\
 &= \sin(x - 90^\circ)\\
 &= -\sin(90^\circ - x)\\
 &= -\cos(x).
\end{align*}
A: hint: $\cos x + \cos (180^{\circ}-x) = 2\cos 90^{\circ}\times \cos(x-90^{\circ}) = 2\times 0 \times \sin x=....?$
A: Notice, the following trigonometric equation $$\color{blue}{\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Hence, if angles $A$ & $B$ are supplementary angles  i.e. $A+B=180^\circ$ $$\implies \cos\left(\frac{A+B}{2}\right)=\cos\left(\frac{180^\circ}{2}\right)=\cos\left(90^\circ\right)=0$$ Hence, we get 
$$\cos(2^\circ)+\cos(178^\circ)=2\cos\left(\frac{2^\circ+178^\circ}{2}\right)\cos\left(\frac{2^\circ-178^\circ}{2}\right)=2\cos(90^\circ)\cos(88^\circ)=0$$
$$\cos(4^\circ)+\cos(176^\circ)=2\cos\left(\frac{4^\circ+176^\circ}{2}\right)\cos\left(\frac{4^\circ-176^\circ}{2}\right)=2\cos(90^\circ)\cos(86^\circ)=0$$
$$\cos(44^\circ)+\cos(136^\circ)=2\cos\left(\frac{44^\circ+136^\circ}{2}\right)\cos\left(\frac{44^\circ-136^\circ}{2}\right)=2\cos(90^\circ)\cos(46^\circ)=0$$
A: Because the two $\cos$ has opposite $x$-values on the unit circle:
A: Using $\cos(\theta)=\cos(-\theta)$ and $\sin(\theta)=-\sin(-\theta)$
$$e^{i\theta}+e^{i(\pi-\theta)}\\
=e^{i\theta}+e^{i\pi}.e^{-i\theta}\\
=\cos(\theta)+i\sin(\theta)-[\cos(-\theta)+i\sin(-\theta)]\\
=cos(\theta)+i\sin(\theta)-[\cos(\theta)-i.\sin(\theta)]\\
=2i\sin(\theta)$$
So the real part of $e^{i\theta}+e^{i(\pi-\theta)}$ is zero, hence $\cos(\theta)+\cos(\pi-\theta)=0$.
