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Trigonometry help!!!!!!!!!!!!!!please??????

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if, cosA+secA =√3, then prove, cos^3A+sec^3A=0
asked Apr 29, 2013 in TRIGONOMETRY by linda Scholar

2 Answers

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cosA + secA = √3

cosA + 1 / cosA = √3

Multiply each side by cosA

cos2A + 1 = √3cosA

cos2A -√3cosA + 1 = 0

Substitute a = 1 , b = -√3 and c = 1 in the quadratic form

cosA = (√3 ± √3 - 4) / 2(1)

          = (√3 ± √ -1) / 2   

But -1 = i2

        =(√3 ± √i2) / 2

cosA=(√3 ± i ) / 2

secA = 1 / cosA = 2 / (√3 ± i)

                             = 2( (√3 ∓ i) / (√3 ∓ i) (√3 ± i))

                             = 2(√3 ∓ i) / (3 - i2)

Substitute i2 = -1

                              = 2(√3 ∓ i) / (3 + 1)

secA                     = (√3 ∓ i) / 2

cos3A + sec3A = [(√3 ± i ) / 2]3 + [(√3 ∓ i) / 2]3

                               = 1 / 8[(√3 ± i )3 + √3 ∓ i)3]

(a ± b)3 + (a ∓ b)3 = 2a(a2 + 3b2) and 2a(a2 + 3b2)

Substitute a = √3 and b = i and a2 = 3 and b2 = -1 in the above formula

                               = 1 / 8[ 2√3(3 + 3(-1)] and 1 / 8[ 2√3(3 + 3(-1)]

Simplify

                               = 0 and 0

                               = 0.

 

answered Apr 29, 2013 by diane Scholar
0 votes

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answered Jul 14, 2014 by david Expert

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