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Trigonometry

In a triangle ABC, the angles A, B, C are in A.P. Show that 2 cos [(A-C)/2] = a+c/sqrt(a^2-ac+c^2)

Apoorva Anand

12 Years agoGrade

2 Answers

Arun Kumar

12 Years ago

Hi

$\\2B=A+C \\A+B+C=180 \\=>B=60 \\b=\sqrt{a^2+c^2-2ac *cosB}=\sqrt{a^2+c^2-ac} \\ {a+c \over b } ={sinA+sinC \over sinB}= {2 sin{(A + C) \over 2} cos{(A - C) \over 2} \over sinB} \\ => {a+c \over b }={2 sin{(2B) \over 2} cos{(A - C) \over 2} \over sinB} \\ => {a+c \over b }={2 sin{B} cos{(A - C) \over 2} \over sinB} \\ => {a+c \over b }=cos{(A - C) \over 2}$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Akshay Chaudhari

12 Years ago

By cosine law i.e. 2accosB=a^2+c^2-b^2

RHS=a+c/b

By sine law i.e. a=2RSinA,and similarly

Further applying Formula of SinA + SinC in LHS

Hence LHS=RHS

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