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Suppose that sin(theta)=15/17; pi/2 less than theta less than pi.

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Find exact value of tan(theta/2)?

asked Mar 21, 2013 in TRIGONOMETRY by angel12 Scholar

2 Answers

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sinθ = 15 / 17 where pi/2 < θ < pi

Recall : Trigonometry formulas sin2(θ) + cos2(θ) =1

              Subtract sin2(θ) from each side

              cos2(θ) = 1 - sin2(θ)

             Take square root each side

             cos(θ) = √(1 - sin2(θ))

Substitute sinθ = 15 / 17 in the equation cos(θ) = √(1 - sin2(θ))

There fore cos(θ) = √(1 - (15/17)2)

cos(θ) = √(1 -225/289)

cos(θ) = √((289  - 225)/289)

cos(θ) = √(64/289)

cos(θ) = -8/17
Trigonometry formulas Tan(θ) = sin(θ) / cos(θ)

Substitute sin(θ) = 15/17 and cos(θ) = 8/17 in the tan(θ) =sin(θ) / cos(θ)

There fore Tan(θ) = (15/17) / (-8/17)

Tan(θ) = -15/8

Trigonometry formulas Tan(θ) = 2Tan(θ/2) / 1 - Tan2(θ/2)

Substitute Tan(θ) = -15/8

-15/8 = 2Tan(θ/2) / 1 - Tan2(θ/2)

Put Tan(θ/2) = t

-15 / 8 = 2t  / 1 - t2

Multiply each side by 8

-15 = 8(2t) / 1 - t2

Multiply each side by 1 - t2

-15(1 - t2) = 16t

Distributive property a(b - c) = ab - ac

-15 + 15t2 = 16t

Subtract 16t  from each side

15t2 -16t -15 = 0

 15t2 -25t +9t - 15 = 0

Common term 5t and 3

5t(3t - 5) +3(3t - 5) =0

Common term 3t - 5

There fore (3t - 5)(5t + 3) = 0

3t - 5 = 0 or 5t + 3 = 0

Case1 :

3t - 5 = 0

Add 5  to each side

3t = 5

Divide each side by 3

t = 5/3

Substitute t = Tan(θ/2)

Tan(θ/2) = 5/3 where  pi/2 < θ <  pi

pi/4 < θ/2 < pi/2

There fore Tan(θ/2) = possitive value

But Tan(θ/2)  = 5 / 3 is a possitive value

Case 1 is a right choice

Case 2 :

5t + 3 = 0

Subtract 3 from each side

5t = -3

Divide each side by 5

t = -3/5

Substitute t = Tan(θ/2)

There fore  Tan(θ/2) = -3/5 this is a possitive value

Case 2 is a wrong choice.

 

           

 

 

answered Mar 21, 2013 by diane Scholar
0 votes

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From the pythagorean theorem image

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But theta lies in image

In second quadrant image is negative.

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From half angle identities image

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In image   ,image is positive.

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answered Jul 1, 2014 by david Expert

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