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

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2 days ago

2074 Points
```							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)^2Let p= cosπ/7cos2π/7cos4π/7so 2psinπ/7= (2sinπ/7cosπ/7)cos2π/7cos4π/7= sin2π/7cos2π/7cos4π/7or 4psinπ/7= (2sin2π/7cos2π/7)cos4π/7= sin4π/7cos4π/7or 8psinπ/7= 2sin4π/7cos4π/7= sin8π/7= – sinπ/7so p= – 1/8.so (a+b)(b+c)(c+a)= – 8(p)^2= – 1/8hence, 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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15 hours ago
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• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions