Each of all the other trigonometric functions of θ in terms of cos θ .

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Answer 1

Question

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

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Answer 2

Question

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

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Answer 3

Question

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

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Answer 4

Question

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

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Answer 5

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

Answer 6

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

Answer 7

Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Answer 8

Expert Solution & Answer

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $θ$ in terms of $cosθ$ are:

$sinθ=±1−cos2θ,tanθ=±1−cos2θcosθ,cscθ=1±1−cos2θ,cotθ=cosθ±1−cos2θ,$ and

$secθ=1cosθ$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Formula:

Reciprocal identities:

$sinu=1cscu,cosu=1secu,tanu=1cotucscu=1sinu,secu=1cosu,cotu=1tanu$ ,

Quotient Identities:

$tanu=sinucosu,cotu=cosusinu$

Pythagorean Identities:

$sin2u+cos2u=11+tan2u=sec2u1+cot2u=csc2u$

Calculation:

Consider the provided trigonometric function $cosθ$ . All the other trigonometric functions has to be expressed in terms of $cosθ$ .

Use the Pythagorean identity $sin2u+cos2u=1$ to calculate $sinθ$ . So,

$sin2θ+cos2θ=1$

Subtract $cos2θ$ from each side of the equation.

$sin2θ+cos2θ−cos2θ=1−cos2θsin2θ=1−cos2θ$

Take square root on both the sides of the equation.

$sin2θ=1−cos2θsinθ=±1−cos2θ$

Now, use the reciprocal identity $cscu=1sinu$ and substitute $sinθ=±1−cos2θ$ to calculate $cscθ$ .

$cscθ=1sinθ=1±1−cos2θ$

By the reciprocal identity $secu=1cosu$ , $secθ$ is calculated in terms of $cosθ$ as,

$secθ=1cosθ$

Use the quotient identity $tanu=sinucosu$ and substitute $sinθ=±1−cos2θ$ to calculate $tanθ$ . So,

$tanθ=±1−cos2θcosθ$

By the reciprocal identity $cotu=1tanu$ , $cotθ$ is calculated in terms of $cosθ$ as,

$cotθ=1tanθ=1±1−cos2θcosθ=±cosθ1−cos2θ$

Thus, all the other trigonometric functions are expressed in terms of $cosθ$ .

Answer 12

Ch. 5.1 - Use the conditions tanx=13 and cosx0 to find the...Ch. 5.1 - Prob. 2CP Ch. 5.1 - Factor each expression. a. 1-cos2 b. 2csc2-7csc+6 Ch. 5.1 - Prob. 4CP Ch. 5.1 - Prob. 5CP Ch. 5.1 - Prob. 6CP Ch. 5.1 - Prob. 7CP Ch. 5.1 - Use the substitution x=3sin,02, to write 9-x2 as a...Ch. 5.1 - Rewrite ln|secx|+ln|sinx| as a single logarithm...Ch. 5.1 - Vocabulary: Fill in the blank to complete the...

Each of all the other trigonometric functions of θ in terms of cos θ .

Concept explainers

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To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .

Answer to Problem 1PS

Explanation of Solution

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Chapter 5 Solutions

Each of all the other trigonometric functions of θ in terms of cos θ .

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To write: Each of all the other trigonometric functions of θ<m:math><m:mi>θ</m:mi></m:math> in terms of cosθ<m:math><m:mrow><m:mi>cos</m:mi><m:mi>θ</m:mi></m:mrow></m:math>.

Answer to Problem 1PS

Explanation of Solution

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Chapter 5 Solutions

To write:

Each of all the other trigonometric functions of $θ$ in terms of $cosθ$ .