Value of $ \cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $?

Question

I am a little weak in trigonometry. I have two questions:

1. Find the value of $\cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $

2. Find the value of $\sin 28^{\circ}+ \cos 17^{\circ} + \cos 28^{\circ} + \sin 17^{\circ} $

I am asking these questions because:
1. I am weak and unable to solve these.
2. I want to know the difference in these questions.

Answer 1

Note that, $$ \cos(\alpha\pm \beta)=\cos \alpha\cos \beta\mp\sin \alpha\sin \beta $$ and $$ \cos(180^\circ-\alpha)=-\cos\alpha. $$ Hence $$ \begin{align} \cos52^\circ+\cos68^\circ+\cos172^\circ&=\color{blue}{\cos(60^\circ-8^\circ)+\cos(60^\circ+8^\circ)}+\color{red}{\cos(180^\circ-8^\circ)}\\ &=\color{blue}{2\cos60^\circ\cos8^\circ}+(\color{red}{-\cos8^\circ})\\ &=\color{blue}{2\cdot\frac12\cdot\cos8^\circ}-\color{red}{\cos8^\circ}\\ &=0. \end{align} $$

Answer 2

Using Prosthaphaeresis Formulas

$$\cos52^\circ+\cos68^\circ=2\cos60^\circ\cos8^\circ$$

Alternatively, $\displaystyle52=60-8,68=60+8\cos(A-B)+\cos(A+B)=?$

and $$\cos172^\circ=\cos(180^\circ-8^\circ)=-\cos8^\circ$$

Answer 3

Observe that $$\cos3(\underbrace{60^\circ-x})=\cos(180^\circ-3x)=-\cos3x$$

and $$\cos3(\underbrace{60^\circ+x})=\cos(180^\circ+3x)=-\cos3x$$

and $$\cos3(\underbrace{180^\circ-x})=\cos(360^\circ+180^\circ-3x)=\cos(180^\circ-3x)=-\cos3x$$

As $\displaystyle\cos3x=4\cos^3x-3\cos x,$

and as $\displaystyle-\cos3x=\cos(180^\circ-3x)=\cos(180^\circ+3x)=\cos3(180^\circ-x)$

the roots of $\displaystyle4\cos^3y-3\cos y=-\cos3x\iff4\cos^3y-3\cos y+\cos3x=0$

will be $\displaystyle\cos(60^\circ-x),\cos(60^\circ+x),\cos(180^\circ-x)$ (no two are in general same, right?)

Using Vieta's formula $\displaystyle\cos(60^\circ-x)+\cos(60^\circ+x)+\cos(180^\circ-x)=\frac04$

Here $\displaystyle x=8^\circ$