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4(cos 2π/15+cos4π/15-cos 7π/15-cos π/15) please tell the solution

4(cos 2π/15+cos4π/15-cos 7π/15-cos π/15) please tell the solution

Grade:12

1 Answers

Arun
25763 Points
3 years ago
π/10 = A....5A = (π/2)....2A = (π/2) - 3A 
sin2A = sin[(π/2) - 3A] 
sin2A = cos3A 
2sinAcosA = 4cos³A - 3cosA ....cosA ≠ 0 
2sinA = 4cos²A - 3 
2sinA = 4(1 - sin²A) - 3 
2sinA = 4 - 4sin²A - 3 
4sin²A + 2sinA - 1 = 0 ....... sinA > 0 
so , sinA = (- 1 +√5)/4 = sin(π/10) 
Next 
cos(π/5) = 1 - 2sin²(π/10) 
= 1 - 2[(- 1 +√5)/4]² 
= 1 - 2[(6 - 2√5)/16] 
= (4/4) - [(3 - √5)/4] 
= (1 + √5)/4 
Then 
cos(2π/15) + cos(4π/15) + cos(8π/15) + cos(16π/15) 
= cos(2π/15) + cos(4π/15) - cos(7π/15) - cos(π/15) 
= [cos(4π/15) + cos(2π/15)] - [cos(7π/15) + cos(π/15)] 
= 2cos(3π/15)cos(π/15) - 2cos(4π/15)cos(3π/15) 
= 2cos(3π/15)[cos(π/15) - cos(4π/15)] 
= 2cos(π/5)*2[sin(π/6)sin(π/10)] 
= 4cos(π/5)sin(π/10)sin(π/6) 
= 4[(1 + √5)/4][(- 1 + √5)/4](1/2) 
= 4(1/4)(1/2) 
= 1/2
But un the question r is multiplied.
Hence answer should be 4 *1/2 = 2

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