Question

If ABCD is a quadrilateral then cos[A/2+B/2]+cos[C/2+D/2]=​

Answer 1

Step-by-step explanation:

see the above attachment

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Answer 2

Answer:

If ABCD is a quadrilateral then cos[A/2+B/2]+cos[C/2+D/2] = 0 .

Step-by-step explanation:

• Let, ABCD be a quadrilateral in which A, B, C, D are interior angles of quadrilateral.
• Sum of interior angles of a quadrilateral is 360⁰ .
• From cosine transformation formula :

$\cos \: c + \cos \: d = 2\cos( \frac{ c+ d}{2} ) \cos( \frac{ c- d}{2} )$

Given that :

• Cos(A/2 + B/2) + cos( C/2 + D/2)

Solution :

• L.H.S = Cos(A/2 + B/2) + cos( C/2 + D/2)

$L.H.S. =\cos(\frac{a}{2} + \frac{b}{2})+\cos( \frac{c}{2} + \frac{d}{2}) \\ =2\cos( \frac{\frac{a}{2} + \frac{b}{2} + \frac{c}{2} + \frac{d}{2}}{2} ) \cos( \frac{\frac{a}{2} + \frac{b}{2} - \frac{c}{2} - \frac{d}{2}}{2} ) \\= 2\cos( \frac{a+b+c+d }{4} )\cos( \frac{a+b-c-d }{4} )$

• Since, A + B + C + D = 360⁰. (A + B + C + D)/4 = 90⁰.
• Cos 90⁰ = 0.

$=0\times\cos( \frac{a+b-c-d }{4} ) \\ = 0$

• Hence, Cos(A/2 + B/2) + cos( C/2 + D/2) = 0.