Question

In any cyclic quadrilateral ABCD cosA + cosB + cosC + cosD=?

Answer 1

a cyclic quadrilateral, is simply a quadrilateral that can be inscribed into a circle,

Answer 2

i knw that opp side sum=180

Answer 3

i want to knw oabt cos

Answer 4

0 i guess

Answer 5

thats what i was thinking A.Avinash_Goutham,
because a square is a cyclic quadrilateral and al the angles are 90°, we know cos(90°)=0

Answer 6

@A.Avinash_Goutham u r correct the answer is 0 how???

Answer 7

perhaps their is a more general solution/

Answer 8

can you use trig identities to prove the general case

Answer 9

Can u show it.........

Answer 10

that's simple...
let a nd c be opp angles
cos (A) + cos(c)=2cos(90)*............... trig identity
(a+b)/2=90
same 4 da other 1

Answer 11

How (a+b)/2=90 ??????

Answer 12

"i knw that opp side sum=180"

Answer 13

yes

Answer 14

i got it

Answer 15

cos (A) + cos(c)=2cos(90)*>??????

Answer 16

\[\cos A + \cos B + \cos C + \cos D\]\[=2\cos\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\right)+2\cos\left(\frac{B+D}{2}\right)\cos\left(\frac{B-D}{2}\right)\]

Answer 17

\[
D = 2\pi -(A+B + C)\\
\cos(D) = \cos(2π−(A+B+C)) = \cos(−(A+B+C))=\cos(A+B+C)
\]
does this help?

Answer 18

\[=2\cos\left(\frac{180}{2}\right)\cos\left(\frac{A-C}{2}\right)+2\cos\left(\frac{180}{2}\right)\]\[=2\cos(90°)\cos\left(\frac{A-C}{2}\right)+2\cos(90°)\cos\left(\frac{B-D}{2}\right)\]\[=0\times\cos\left(\frac{A-C}{2}\right)+0\times\cos\left(\frac{B-D}{2}\right)\]\[=0\]

Answer 19

@UnkleRhaukus is there a formula

Answer 20

in the first step i used this sum-to-product trig formula
\[\cos(x)+\cos(y)=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\]

Answer 21

now got it thanzzz @UnkleRhaukus

Answer 22

u explained so well

Answer 23

we dont know the difference of opposite angles (x-y) but we do know or at least you knew that the sum of opposite angles (x+y=180)

Answer 24

\[
A= \pi - D\\
B= \pi -C\\
\cos(A) = \cos(\pi -D)= -\cos ( D)\\
\cos(B) =\cos(\pi -C)= -\cos ( C)\\
\cos(A) + \cos(B)+ \cos(C) + \cos(D)=\\
-\cos(D) - \cos(C)+ \cos(C) + \cos(D)=0\
\]

Answer 25

I used that in a cyclic quadrilateral opposite angles sum up to to 180 degrees.

Answer 26

that is good proof too @eliassaab

Answer 27

Thanks @UnkleRhaukus