High school Trigonometry

We can see that the angles are in Arithmetic Progression,

we know,

$\cos\alpha + cos(\alpha + \beta) + cos(\alpha + 2\beta) + ... + \cos(\alpha + (n-1)\beta)$

$= \frac{\cos(\alpha+ \frac{n-1}{2}\beta)\sin(\frac{n\beta}{2})}{sin(\frac{\beta}{2})}$

So, this expression would give

$= \frac{\cos159^{\circ}\sin180^{\circ}}{\sin36^{\circ}}$

$\sin180^{\circ} = 0$

Therefore the answer is $0$

$\cos{15^\circ} + \cos{87^\circ} \color{#D61F06}{+ \cos{159^\circ}} \color{#3D99F6}{+ \cos{231^\circ}} \color{#20A900}{+\cos{303^\circ}} \\ = \cos{15^\circ} + \cos{87^\circ} \color{#D61F06}{- \cos{21^\circ}} \color{#3D99F6}{- \cos{51^\circ}} \color{#20A900} {+ \cos{57^\circ}} \\ = \cos{15^\circ} - \cos{21^\circ} - \cos{51^\circ} + \cos{57^\circ} + \cos{87^\circ} \\ = \color{#D61F06}{\cos{(18-3)^\circ} - \cos{(18+3)^\circ}} - \color{#3D99F6}{\cos{(54-3)^\circ} + \cos{(54+3)^\circ}} + \color{#20A900} {\cos{87^\circ}} \\ = \color{#D61F06}{2\sin{18^\circ} \sin{3^\circ}} + \color{#3D99F6}{2\sin{54^\circ} \sin{3^\circ}} + \color{#20A900}{\sin{3^\circ}} \\ = 2\sin{3^\circ} \left(\color{#D61F06}{\sin{18^\circ}} - \color{#3D99F6}{\sin{54^\circ}} + \color{#20A900}{\frac{1}{2}} \right) \\ = 2\sin{3^\circ} \left( \color{#D61F06}{\cos{72^\circ}} \color{#3D99F6}{- \cos{36^\circ}} + \frac{1}{2} \right) \\ = 2\sin{3^\circ} \left( \left[ \cos{72^\circ} \color{#3D99F6}{+ \cos{144^\circ}} \right] + \frac{1}{2} \right) \\ = 2\sin{3^\circ} \left( \left[-\frac{1}{2}\right] + \frac{1}{2} \right) = \boxed{0}$

$\cos15^{\circ}+\cos87^{\circ}+\cos159^{\circ}+\cos231^{\circ}+\cos303^{\circ}$

$=\cos15^{\circ}+\cos87^{\circ}-\cos21^{\circ}-\cos51^{\circ}+\cos57^{\circ}$

$=(\cos15^{\circ}-\cos21^{\circ})-(\cos51^{\circ}-\cos57^{\circ})+\cos87^{\circ}$

$=2\sin18^{\circ}\sin3^{\circ}-2\sin54^{\circ}\sin3^{\circ}+\cos87^{\circ}$

$=2\sin3^{\circ}(\sin18^{\circ}-\sin54^{\circ})+\cos87^{\circ}$

$=2\sin3^{\circ}(-2\cos36^{\circ}\sin18^{\circ})+\cos87^{\circ}$

$=2\sin3^{\circ}(-\frac{2\cos36^{\circ}\sin18^{\circ}\cos18^{\circ}}{\cos18^{\circ}})+\cos87^{\circ}$

$=2\sin3^{\circ}(-\frac{\sin36^{\circ}\cos36^{\circ}}{\cos18^{\circ}})+\cos87^{\circ}$

$=2\sin3^{\circ}(-\frac{0.5\sin72^{\circ}}{\cos18^{\circ}})+\cos87^{\circ}$

$=2\sin3^{\circ}\cdot (-0.5)+\cos87^{\circ}$

$=-\sin3^{\circ}+\cos87^{\circ}$

$=\boxed{0}$ .

$cos 15^{\circ} + cos (180 - 21)^{\circ} + cos (180 + 51)^{\circ} + cos (360 - 57)^{\circ} + cos 87^{\circ}$ $= cos 15^{\circ} - cos 21^{\circ} - cos 51^{\circ} + cos 57^{\circ} + cos 87^{\circ}$ $= 2 sin 18^{\circ} sin 3^{\circ} - 2 sin 54^{\circ} sin 3^{\circ} + cos 87^{\circ}$ $= - sin 3^{\circ} + cos 87^{\circ} = 0$ , using well known values of $sin 18^{\circ}\ \ and\ \ sin 54^{\circ}.$

Consider the complex nos with moduli 1 and arguments 15, 87, 154, 231 and 303. Since the arguments have a common difference of 72 the complex nos are the fifth roots of unity each multiplied by the complex no with modulus 1 and argument 15. Take out this last complex no as a common factor. The sum of the roots of unity is zero multiplied by the common factor is still zero. The sum of the cosines represents the real part and the real part of zero is still zero. Hence the answer.