Query assignment 5

course mth 164

}`\Zzassignment #006

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006.

Precalculus II

06-15-2007

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19:30:02

Query problem 6.5.10 exact value of sin^-1( -`sqrt(3)/2)

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RESPONSE -->

confidence assessment:

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19:30:15

Query problem 6.5.10 exact value of sin^-1( -`sqrt(3)/2)

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RESPONSE -->

-pi/3

confidence assessment: 3

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assignment #006

006.

Precalculus II

06-15-2007

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19:31:04

Query problem 6.5.10 exact value of sin^-1( -`sqrt(3)/2)

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RESPONSE -->

The answer would be -pi/3

confidence assessment: 3

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19:31:18

what is the exact value of the given expression?

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RESPONSE -->

The exact value is -pi/3

confidence assessment: 3

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19:33:42

Describe the triangle you would use to evaluate this expression, and explain how you would use it.

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RESPONSE -->

I would use a 30, 60, 90 triangle to solve this problem

The 60 degree angle would be at the origin, hypotenuse 1 and drop a perpendicular to the x-axis to form a right triangle

confidence assessment: 2

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19:35:05

Query problem 6.5.34 exact value of cos(sin^-1(-`sqrt(3 / 2 ))?

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RESPONSE -->

since in the previous problem it was found that sin^1(-sqrt(3)/2)) is -pi/3 then we would find

cos(-pi/3) which is equal to 1/2

confidence assessment: 3

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19:35:16

what is the exact value of the given expression?

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RESPONSE -->

The exact value is 1/2

confidence assessment: 3

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19:36:27

Describe the triangle you would use to evaluate this expression, and explain how you would use it.

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RESPONSE -->

I would use the same triangle as before to find the value of sin^1(-sqrt(3)/2)) and then the cos of that angle would be the x coordinat of the vertex of the triangle on the unit circle

confidence assessment: 2

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19:37:17

Query problem 6.5.88 cos(tan^-1(v)) = 1 / `sqrt(1+v^2)

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RESPONSE -->

ok

confidence assessment: 3

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19:39:13

explain how you establish the given identity.

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RESPONSE -->

First set up a right triangle and identify one angle as theta. Since tangent is opposite/adjacent then the opposite leg would be v and the adjacent leg would be 1. Using the pythagorean theorem the hypotenuse would be sqrt(1+v^2). Using the trig ratio for cosine which is adjacent over hypotenuse we find that the cosine of that angle would be 1/sqrt(1+v^2)

confidence assessment: 3

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19:39:27

Describe the triangle you might use to establish this identity and how you would use it.

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RESPONSE -->

I described it in the answer before.

confidence assessment: 3

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19:40:20

Comm on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

Right triangles can be used for a wide range of purposes not just for finding the length of sides from two sides but also from trig ratios not even knowing the measure of the angle.

confidence assessment: 3

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Excellent. Let me know if you have questions.