If A, B, C, D are the angles of a cyclic quadrilateral prove that: cos A + cos B + - Maths - Trigonometric Functions

Question

If A, B, C, D are the angles of a cyclic quadrilateral prove
that: cos A + cos B + cos C + cos D= 0

Santosh Kumar · Accepted Answer

In a cyclic quadrilateral, sum of opposite angles is 180^o.

$A+ C = 180 ° B+D = 180 °$

L.H.S = cos A + cos B + cos C + cos D

$= (\cos A+ cos C)+ (cosB + \cos D) = 2 \cos (\frac{A+C}{2}) . \cos (\frac{A - C}{2}) + 2 \cos (\frac{B+D}{2}) . \cos (\frac{B - D}{2}) = 2 \cos (\frac{180 °}{2}) . \cos (\frac{A - C}{2}) + 2 \cos (\frac{180 °}{2}) . \cos (\frac{B - D}{2}) = 2 cos 90 ° . \cos (\frac{A - C}{2}) + 2 \cos 90 ° . \cos (\frac{B - D}{2}) = 0 [∵ \cos 90 ° = 0] = R.H.S.$

If A, B, C, D are the angles of a cyclic quadrilateral prove that: cos A + cos B + cos C + cos D= 0.