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High School Math Algebra Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

the values of other five trigonometric functions of θ . sin θ = 5 29 29 , csc θ = 29 5 , cos θ = − 2 29 29 , sec θ = − 29 2 , tan θ = − 5 2 . Given: cot θ = − 2 5 , π 2 < θ < π . Concept used: Pythagorean identity: 1 + cot 2 θ = csc 2 θ sin 2 θ + cos 2 θ = 1 Reciprocal identities: csc θ = 1 sin θ , sec θ = 1 cos θ Tangent and Cotangent identities: tan θ = sin θ cos θ , cot θ = cos θ sin θ Calculation: By using the Pythagorean identity 1 + cot 2 θ = csc 2 θ , it follows that 1 + − 2 5 2 = csc 2 θ 1 + 4 25 = csc 2 θ 29 25 = csc 2 θ csc θ = ± 29 5 Now since θ is in quadrant II, csc θ is negative. ∴ csc θ = 29 5 Now by using the reciprocal identity, it follows that sin θ = 1 csc θ = 1 29 5 = 5 29 = 5 29 29 Now by using the Pythagorean identity sin 2 θ + cos 2 θ = 1 , it follows that 5 29 29 2 + cos 2 θ = 1 cos 2 θ = 1 − 725 841 cos 2 θ = 116 841 cos θ = ± 2 29 29 Now since θ is in quadrant II, cos θ is negative. ∴ cos θ = − 2 29 29 Now by using the reciprocal identity, it follows that sec θ = 1 cos θ = 1 − 2 29 29 = − 29 2 29 = − 29 2 Now by using the tangent and identity, it follows that tan θ = 5 29 29 − 2 29 29 = − 5 2

the values of other five trigonometric functions of θ . sin θ = 5 29 29 , csc θ = 29 5 , cos θ = − 2 29 29 , sec θ = − 29 2 , tan θ = − 5 2 . Given: cot θ = − 2 5 , π 2 < θ < π . Concept used: Pythagorean identity: 1 + cot 2 θ = csc 2 θ sin 2 θ + cos 2 θ = 1 Reciprocal identities: csc θ = 1 sin θ , sec θ = 1 cos θ Tangent and Cotangent identities: tan θ = sin θ cos θ , cot θ = cos θ sin θ Calculation: By using the Pythagorean identity 1 + cot 2 θ = csc 2 θ , it follows that 1 + − 2 5 2 = csc 2 θ 1 + 4 25 = csc 2 θ 29 25 = csc 2 θ csc θ = ± 29 5 Now since θ is in quadrant II, csc θ is negative. ∴ csc θ = 29 5 Now by using the reciprocal identity, it follows that sin θ = 1 csc θ = 1 29 5 = 5 29 = 5 29 29 Now by using the Pythagorean identity sin 2 θ + cos 2 θ = 1 , it follows that 5 29 29 2 + cos 2 θ = 1 cos 2 θ = 1 − 725 841 cos 2 θ = 116 841 cos θ = ± 2 29 29 Now since θ is in quadrant II, cos θ is negative. ∴ cos θ = − 2 29 29 Now by using the reciprocal identity, it follows that sec θ = 1 cos θ = 1 − 2 29 29 = − 29 2 29 = − 29 2 Now by using the tangent and identity, it follows that tan θ = 5 29 29 − 2 29 29 = − 5 2

Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

2nd Edition

ISBN: 9780618912407

Author: MCDOUGAL LITTEL

Publisher: Mcdougal Littel

See similar textbooks

Expert Solution & Answer

Chapter 10.3, Problem 7E

Solution

To find: the values of other five trigonometric functions of θ .

sinθ=52929, cscθ=295, cosθ=−22929, secθ=−292, tanθ=−52 .

Given:

cotθ=−25, π2<θ<π .

Concept used:

Pythagorean identity:

1+cot2θ=csc2θsin2θ+cos2θ=1

Reciprocal identities:

cscθ=1sinθ, secθ=1cosθ

Tangent and Cotangent identities:

tanθ=sinθcosθ, cotθ=cosθsinθ

Calculation:

By using the Pythagorean identity 1+cot2θ=csc2θ , it follows that

1+−252=csc2θ1+425=csc2θ2925=csc2θcscθ=±295

Now since θ is in quadrant II, cscθ is negative.

∴cscθ=295

Now by using the reciprocal identity, it follows that

sinθ=1cscθ=1295=529=52929

Now by using the Pythagorean identity sin2θ+cos2θ=1 , it follows that

529292+cos2θ=1cos2θ=1−725841cos2θ=116841cosθ=±22929

Now since θ is in quadrant II, cosθ is negative.

∴cosθ=−22929

Now by using the reciprocal identity, it follows that

secθ=1cosθ=1−22929=−29229=−292

Now by using the tangent and identity, it follows that

tanθ=52929−22929=−52

Chapter 10.3, Problem 6E

Chapter 10.3, Problem 8E

Chapter 10 Solutions

Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

Ch. 10.1 - Prob. 1GP Ch. 10.1 - Prob. 2GP Ch. 10.1 - Prob. 3GP Ch. 10.1 - Prob. 4GP Ch. 10.1 - Prob. 5GP Ch. 10.1 - Prob. 6GP Ch. 10.1 - Prob. 7GP Ch. 10.1 - Prob. 8GP Ch. 10.1 - Prob. 9GP Ch. 10.1 - Prob. 10GP

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