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High School Math Algebra Holt Mcdougal Larson Algebra 2: Student Edition 2012

The values of trigonometric functions cos θ , tan θ , csc θ , sec θ & cot θ . The trigonometric functions are, sin θ = − 11 6 , tan θ = − 11 5 , csc θ = − 6 11 11 , sec θ = 6 5 , cot θ = − 5 11 11 . Given information: The given trigonometric function is cos θ = 5 6 and 3 π 2 < θ < 2 π . Concept and Formula Used: Use the trigonometric formula , sin 2 θ + cos 2 θ = 1 , tan θ = sin θ cos θ , csc θ = 1 sin θ , sec θ = 1 cos θ , cot θ = 1 tan θ . Calculation: The given trigonometric function is cos θ = 5 6 . First find sin θ , Using, sin 2 θ + cos 2 θ = 1 sin 2 θ + 5 6 2 = 1 sin 2 θ + 25 36 = 1 sin 2 θ = 1 − 25 36 sin 2 θ = 36 − 25 36 sin 2 θ = 11 36 sin θ = 11 36 sin θ = ± 11 6 As, the θ lies in fourth quadrant therefore, sin θ = − 11 6 Now, find the other trigonometric functions, tan θ = sin θ cos θ tan θ = − 11 6 5 6 tan θ = − 11 6 × 6 5 tan θ = − 11 5 Now, csc θ = 1 sin θ csc θ = − 6 11 csc θ = − 6 11 × 11 11 csc θ = − 6 11 11 sec θ = 1 cos θ sec θ = 6 5 cot θ = 1 tan θ cot θ = − 5 11 × 11 11 cot θ = − 5 11 11 Conclusion: The trigonometric functions are , sin θ = − 11 6 , tan θ = − 11 5 , csc θ = − 6 11 11 , sec θ = 6 5 , cot θ = − 5 11 11 .

The values of trigonometric functions cos θ , tan θ , csc θ , sec θ & cot θ . The trigonometric functions are, sin θ = − 11 6 , tan θ = − 11 5 , csc θ = − 6 11 11 , sec θ = 6 5 , cot θ = − 5 11 11 . Given information: The given trigonometric function is cos θ = 5 6 and 3 π 2 < θ < 2 π . Concept and Formula Used: Use the trigonometric formula , sin 2 θ + cos 2 θ = 1 , tan θ = sin θ cos θ , csc θ = 1 sin θ , sec θ = 1 cos θ , cot θ = 1 tan θ . Calculation: The given trigonometric function is cos θ = 5 6 . First find sin θ , Using, sin 2 θ + cos 2 θ = 1 sin 2 θ + 5 6 2 = 1 sin 2 θ + 25 36 = 1 sin 2 θ = 1 − 25 36 sin 2 θ = 36 − 25 36 sin 2 θ = 11 36 sin θ = 11 36 sin θ = ± 11 6 As, the θ lies in fourth quadrant therefore, sin θ = − 11 6 Now, find the other trigonometric functions, tan θ = sin θ cos θ tan θ = − 11 6 5 6 tan θ = − 11 6 × 6 5 tan θ = − 11 5 Now, csc θ = 1 sin θ csc θ = − 6 11 csc θ = − 6 11 × 11 11 csc θ = − 6 11 11 sec θ = 1 cos θ sec θ = 6 5 cot θ = 1 tan θ cot θ = − 5 11 × 11 11 cot θ = − 5 11 11 Conclusion: The trigonometric functions are , sin θ = − 11 6 , tan θ = − 11 5 , csc θ = − 6 11 11 , sec θ = 6 5 , cot θ = − 5 11 11 .

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Holt Mcdougal Larson Algebra 2: Student Edition 2012

1st Edition

ISBN: 9780547647159

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

See similar textbooks

Expert Solution & Answer

Chapter 10.3, Problem 5E

Solution

To Find: The values of trigonometric functions cosθ,tanθ,cscθ,secθ&cotθ .

The trigonometric functions are, sinθ=−116,tanθ=−115,cscθ=−61111,secθ=65,cotθ=−51111 .

Given information:

The given trigonometric function is cosθ=56 and 3π2<θ<2π .

Concept and Formula Used:

Use the trigonometric formula , sin2θ+cos2θ=1,tanθ=sinθcosθ,cscθ=1sinθ,secθ=1cosθ,cotθ=1tanθ .

Calculation:

The given trigonometric function is cosθ=56 .

First find sinθ ,

Using,

sin2θ+cos2θ=1sin2θ+562=1sin2θ+2536=1sin2θ=1−2536sin2θ=36−2536sin2θ=1136sinθ=1136sinθ=±116

As, the θ lies in fourth quadrant therefore,

sinθ=−116

Now, find the other trigonometric functions,

tanθ=sinθcosθtanθ=−11656tanθ=−116×65tanθ=−115

Now,

cscθ=1sinθcscθ=−611cscθ=−611×1111cscθ=−61111secθ=1cosθsecθ=65cotθ=1tanθcotθ=−511×1111cotθ=−51111

Conclusion:

The trigonometric functions are , sinθ=−116,tanθ=−115,cscθ=−61111,secθ=65,cotθ=−51111 .

Chapter 10.3, Problem 4E

Chapter 10.3, Problem 6E

Chapter 10 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

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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