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If 1 = tan(θ)tan(2θ), what are the values of θ?

0 votes

Let's say that I have the expression 1 = tan(θ)tan(2θ). Algebraically, how do I find the repeating values of θ?

I'm not looking for an approximation method. How would this be done algebraically?

Thanks.

asked Jul 13, 2013 in TRIGONOMETRY by harvy0496 Apprentice

2 Answers

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The given expression is 1 = tan(θ)tan(2θ)

1 = tan(θ) * [ 2 tan(θ) / (1 − tan²(θ)) ]
Then multiply both sides by  (1 − tan²(θ)) in the above equation.

(1 − tan²(θ)) = 2 tan²(θ)

1 = 2 tan²(θ) + tan²(θ)
1 = 3 tan²(θ)
1/3 = tan²(θ
)
so  tan(θ) = 1/sqrt(3) (or) tan(θ) = -1/sqrt(3)

tan(θ) = sqrt(3)/3 (or) tan(θ) = -sqrt(3)/3
 

answered Jul 13, 2013 by anonymous

tan(θ) = 1/sqrt(3) (or) tan(θ) = -1/sqrt(3)

tan(θ) = tan(π/6) or tan(θ) = tan(-π/6)

tan(θ) = tan(π/6) or tan(θ) = tan(π - π/6)

tan(θ) = tan(π/6) or tan(θ) = tan(5π/6)

Apply inverse tangent to each side.

θ = nπ + π/6 or θ = nπ + 5π/6

The general solutions are θ = nπ + π/6 or θ = nπ + 5π/6

 

0 votes

The expression is 1 = tan(θ)tan(2θ).

tan(θ) tan(2θ) = 1

[sin(θ) / cos(θ)] [sin(2θ) / cos(2θ)] = 1

sin(θ)sin(2θ) / cos(θ)cos(2θ) = 1

sin(θ){2sin(θ)cos(θ)} / cos(θ)cos(2θ) = 1                 [Since sin(2θ) = 2sin(θ)cos(θ)]

2sin2(θ)cos(θ) / cos(θ)cos(2θ) = 1

2sin2(θ) / cos(2θ) = 1

{1 - cos(2θ)} / cos(2θ) = 1                                  [Since cos(2θ) = 1 - 2sin2(θ)]

1/cos(2θ) - cos(2θ)/cos(2θ) = 1

1/cos(2θ) - 1 = 1

1/cos(2θ) = 1 + 1

1/cos(2θ) = 2

1/2 = cos(2θ)

cos(2θ) = 1/2

cos(2θ) = cos(π / 3)

Take inverse cosine to each side.

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer.

2θ = 2πn₁ ± π / 3 for n₁ Є Z or 2θ = 2πn₂ ± 5π / 3 for n₂ Є Z.

θ = 1/2 (2πn₁ ± π / 3 for n₁ Є Z) or θ = 1/2 (2πn₂ ± 5π / 3) for n₂ Є Z.

The general solutions are θ = πn₁ ± π / 6 for n₁ Є Z or θ = πn₂ ± 5π / 6 for n₂ Є Z.

 

answered Jul 3, 2014 by joly Scholar
edited Jul 3, 2014 by joly

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