Given the triangle ABC, what is cos(A/2)*cos(B/2)*cos(C/2)?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since we'll have to evaluate the product of cosines, we'll apply the law of cosines:

cos A/2= sqrt (p(p-a)/b*c),

Half perimeter p: p=(a+b+c)/2 and a,b,c, are the length of the opposite sides to the A,B,C, angles.

cos B/2= sqrt (p(p-b)/a*c)

cos C/2= sqrt (p(p-c)/b*a)

We'll evaluate the product:

cos A/2*cos B/2*cos C/2= sqrt [p(p-a)*p(p-b)*p(p-c)/a^2*b^2*c^2]

cos A/2*cos B/2*cos C/2=p*sqrt[p(p-a)*(p-b)*(p-c)]/a*b*c

We recognize the Heron's formula for evaluating the area:

A = sqrt[p(p-a)*(p-b)*(p-c)]

cos A/2*cos B/2*cos C/2=p*A/a*b*c

The requested product is: cos A/2*cos B/2*cos C/2=p*A/a*b*c

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