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prove that: cos^3 (π/7)+cos^3 (3π/7)+cos^3 (5π/7) = 1/2

prove that:
cos^3 (π/7)+cos^3 (3π/7)+cos^3 (5π/7) = 1/2

Grade:10

1 Answers

Aditya Gupta
2081 Points
2 years ago
Use the identity (a+b+c)^3= a^3+b^3+c^3 + 3(a+b)(b+c)(c+a) where a b and c are cos(π/7), cos(3π/7), cos(5π/7) respectively.
note that a+b+c= cos(π/7+2π/7)*sin(3π/7)/sin(π/7)  (as π/7, 3π/7, 5π/7 are in AP, we have used std formula)
= 2cos(3π/7)sin(3π/7)/2sin(π/7)= sin6π/7/2sinπ/7= ½ as sin6π/7= sinπ/7.
a+b= 2cos2π/7*cosπ/7, b+c= 2cos4π/7cosπ/7, c+a= 2cos3π/7*cos2π/7= – 2cos4π/7cos2π/7 as cos3π/7= – cos4π/7.
so, (a+b)(b+c)(c+a)= – 8(cosπ/7cos2π/7cos4π/7)^2
Let p= cosπ/7cos2π/7cos4π/7
so 2psinπ/7= (2sinπ/7cosπ/7)cos2π/7cos4π/7= sin2π/7cos2π/7cos4π/7
or 4psinπ/7= (2sin2π/7cos2π/7)cos4π/7= sin4π/7cos4π/7
or 8psinπ/7= 2sin4π/7cos4π/7= sin8π/7= – sinπ/7
so p= – 1/8.
so (a+b)(b+c)(c+a)= – 8(p)^2= – 1/8
hence, a^3+b^3+c^3= (1/2)^3 – 3*( – 1/8) = 1/8 + 3/8= ½ 
hence proved.
KINDLY APPROVE :D 

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