Basic trigo - IV

Since it is a cyclic quadrilateral, opposite angles are supplementary.

So, $A+C={ 180 }^{ \circ }\quad$ and $B+D={ 180 }^{ \circ }\quad$

$\Rightarrow C={ 180 }^{ \circ }-A$ and $D={ 180 }^{ \circ }-B$

so, $\cos { A } +\cos { B } +\cos { C } +\cos { D=\quad \cos { A } +\cos { B } +\cos { \left( { 180 }^{ \circ }-A \right) } +\cos { \left( { 180 }^{ \circ }-B \right) } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \cos { A } +\cos { B } -\cos { A } -\cos { B=\quad 0 }$

since it is a cyclic quadrilateral Opposite angles are supplementary

so A + C =180 and B + D =180 degrees

CosA + CosB + CosC + CosD can be written as

(CosA + CosC) + (CosB + CosD)

applying Cosa + cosb formulae from trignometric functions

2Cos(A+C)/2 Cos(A-C)/2 + 2Cos(B+D)/2 Cos(B-D)/2 since A + C =180 B + D = 180

(A + C)/2 = 90 degrees and (B + D)/2 = 90 since Cos 90 = 0

Then the whole above equation minimizes to Zero!!

Hope this helps.....

The identity of cos + cosB+ cosC+ cosD = 1-4sinA/2sinB/2sinC/2sinD/2 may also be used

Assume $A$ = $B$ = $C$ = $D$ = $90$

$4\cos{A} = 4\cos{90} = \boxed{0}$

Is it a correct solution? haha