4(cos 2π/15+cos4π/15-cos 7π/15-cos π/15) please tell the solution
Dhanush , 6 Years ago
Grade 12
Trigonometry> 4(cos 2π/15+cos4π/15-cos 7π/15-cos π/15) ...
1 Answers
Arun
Last Activity: 6 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
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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