Find The Five Remaining Trigonometric Functions Of { \theta$}$ Using The Given Information:Given:$ \cos \theta = -\frac{1}{4}, \quad \sin \theta \ \textgreater \ 0 }$Complete The Following $[ \begin{array {ll} \sin \theta =

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Introduction

In this article, we will explore the process of finding the five remaining trigonometric functions of {\theta$}$ given the information that cos⁡θ=−14{\cos \theta = -\frac{1}{4}} and sin⁡θ>0{\sin \theta > 0}. We will use the given information to find the values of the remaining trigonometric functions, including sin⁡θ{\sin \theta}, tan⁡θ{\tan \theta}, csc⁡θ{\csc \theta}, sec⁡θ{\sec \theta}, and cot⁡θ{\cot \theta}.

Recall of Trigonometric Identities

Before we begin, let's recall some important trigonometric identities that we will use to find the remaining trigonometric functions.

  • Pythagorean Identity: sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1}
  • Reciprocal Identities: csc⁡θ=1sin⁡θ{\csc \theta = \frac{1}{\sin \theta}}, sec⁡θ=1cos⁡θ{\sec \theta = \frac{1}{\cos \theta}}, and cot⁡θ=1tan⁡θ{\cot \theta = \frac{1}{\tan \theta}}
  • Quotient Identity: tan⁡θ=sin⁡θcos⁡θ{\tan \theta = \frac{\sin \theta}{\cos \theta}}

Finding sin⁡θ{\sin \theta}

Given that cos⁡θ=−14{\cos \theta = -\frac{1}{4}} and sin⁡θ>0{\sin \theta > 0}, we can use the Pythagorean identity to find the value of sin⁡θ{\sin \theta}.

sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1}

Substituting the given value of cos⁡θ{\cos \theta}, we get:

sin⁡2θ+(−14)2=1{\sin^2 \theta + \left(-\frac{1}{4}\right)^2 = 1}

Simplifying the equation, we get:

sin⁡2θ+116=1{\sin^2 \theta + \frac{1}{16} = 1}

Subtracting 116{\frac{1}{16}} from both sides, we get:

sin⁡2θ=1516{\sin^2 \theta = \frac{15}{16}}

Taking the square root of both sides, we get:

sin⁡θ=±1516{\sin \theta = \pm \sqrt{\frac{15}{16}}}

Since we are given that sin⁡θ>0{\sin \theta > 0}, we take the positive square root:

sin⁡θ=1516=154{\sin \theta = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}}

Finding tan⁡θ{\tan \theta}

Now that we have found the value of sin⁡θ{\sin \theta}, we can use the quotient identity to find the value of tan⁡θ{\tan \theta}.

tan⁡θ=sin⁡θcos⁡θ{\tan \theta = \frac{\sin \theta}{\cos \theta}}

Substituting the values of sin⁡θ{\sin \theta} and cos⁡θ{\cos \theta}, we get:

tan⁡θ=154−14{\tan \theta = \frac{\frac{\sqrt{15}}{4}}{-\frac{1}{4}}}

Simplifying the expression, we get:

tan⁡θ=−15{\tan \theta = -\sqrt{15}}

Finding csc⁡θ{\csc \theta}

Now that we have found the value of sin⁡θ{\sin \theta}, we can use the reciprocal identity to find the value of csc⁡θ{\csc \theta}.

csc⁡θ=1sin⁡θ{\csc \theta = \frac{1}{\sin \theta}}

Substituting the value of sin⁡θ{\sin \theta}, we get:

csc⁡θ=1154{\csc \theta = \frac{1}{\frac{\sqrt{15}}{4}}}

Simplifying the expression, we get:

csc⁡θ=415{\csc \theta = \frac{4}{\sqrt{15}}}

Rationalizing the denominator, we get:

csc⁡θ=41515{\csc \theta = \frac{4\sqrt{15}}{15}}

Finding sec⁡θ{\sec \theta}

Now that we have found the value of cos⁡θ{\cos \theta}, we can use the reciprocal identity to find the value of sec⁡θ{\sec \theta}.

sec⁡θ=1cos⁡θ{\sec \theta = \frac{1}{\cos \theta}}

Substituting the value of cos⁡θ{\cos \theta}, we get:

sec⁡θ=1−14{\sec \theta = \frac{1}{-\frac{1}{4}}}

Simplifying the expression, we get:

sec⁡θ=−4{\sec \theta = -4}

Finding cot⁡θ{\cot \theta}

Now that we have found the value of tan⁡θ{\tan \theta}, we can use the reciprocal identity to find the value of cot⁡θ{\cot \theta}.

cot⁡θ=1tan⁡θ{\cot \theta = \frac{1}{\tan \theta}}

Substituting the value of tan⁡θ{\tan \theta}, we get:

cot⁡θ=1−15{\cot \theta = \frac{1}{-\sqrt{15}}}

Simplifying the expression, we get:

cot⁡θ=−115{\cot \theta = -\frac{1}{\sqrt{15}}}

Rationalizing the denominator, we get:

cot⁡θ=−1515{\cot \theta = -\frac{\sqrt{15}}{15}}

Conclusion

In this article, we have found the five remaining trigonometric functions of {\theta$}$ given the information that cos⁡θ=−14{\cos \theta = -\frac{1}{4}} and sin⁡θ>0{\sin \theta > 0}. We have used the Pythagorean identity, reciprocal identities, and quotient identity to find the values of sin⁡θ{\sin \theta}, tan⁡θ{\tan \theta}, csc⁡θ{\csc \theta}, sec⁡θ{\sec \theta}, and cot⁡θ{\cot \theta}. The values of the trigonometric functions are:

  • sin⁡θ=154{\sin \theta = \frac{\sqrt{15}}{4}}
  • tan⁡θ=−15{\tan \theta = -\sqrt{15}}
  • csc⁡θ=41515{\csc \theta = \frac{4\sqrt{15}}{15}}
  • sec⁡θ=−4{\sec \theta = -4}
  • cot⁡θ=−1515{\cot \theta = -\frac{\sqrt{15}}{15}}

Q: What are the five remaining trigonometric functions of {\theta$}$ given the information that cos⁡θ=−14{\cos \theta = -\frac{1}{4}} and sin⁡θ>0{\sin \theta > 0}?

A: The five remaining trigonometric functions of {\theta$}$ are sin⁡θ{\sin \theta}, tan⁡θ{\tan \theta}, csc⁡θ{\csc \theta}, sec⁡θ{\sec \theta}, and cot⁡θ{\cot \theta}.

Q: How do I find the value of sin⁡θ{\sin \theta} given that cos⁡θ=−14{\cos \theta = -\frac{1}{4}} and sin⁡θ>0{\sin \theta > 0}?

A: To find the value of sin⁡θ{\sin \theta}, you can use the Pythagorean identity: sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1}. Substituting the given value of cos⁡θ{\cos \theta}, you get: sin⁡2θ+(−14)2=1{\sin^2 \theta + \left(-\frac{1}{4}\right)^2 = 1}. Simplifying the equation, you get: sin⁡2θ+116=1{\sin^2 \theta + \frac{1}{16} = 1}. Subtracting 116{\frac{1}{16}} from both sides, you get: sin⁡2θ=1516{\sin^2 \theta = \frac{15}{16}}. Taking the square root of both sides, you get: sin⁡θ=±1516{\sin \theta = \pm \sqrt{\frac{15}{16}}}. Since you are given that sin⁡θ>0{\sin \theta > 0}, you take the positive square root: sin⁡θ=1516=154{\sin \theta = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}}.

Q: How do I find the value of tan⁡θ{\tan \theta} given that sin⁡θ=154{\sin \theta = \frac{\sqrt{15}}{4}} and cos⁡θ=−14{\cos \theta = -\frac{1}{4}}?

A: To find the value of tan⁡θ{\tan \theta}, you can use the quotient identity: tan⁡θ=sin⁡θcos⁡θ{\tan \theta = \frac{\sin \theta}{\cos \theta}}. Substituting the values of sin⁡θ{\sin \theta} and cos⁡θ{\cos \theta}, you get: tan⁡θ=154−14{\tan \theta = \frac{\frac{\sqrt{15}}{4}}{-\frac{1}{4}}}. Simplifying the expression, you get: tan⁡θ=−15{\tan \theta = -\sqrt{15}}.

Q: How do I find the value of csc⁡θ{\csc \theta} given that sin⁡θ=154{\sin \theta = \frac{\sqrt{15}}{4}}?

A: To find the value of csc⁡θ{\csc \theta}, you can use the reciprocal identity: csc⁡θ=1sin⁡θ{\csc \theta = \frac{1}{\sin \theta}}. Substituting the value of sin⁡θ{\sin \theta}, you get: csc⁡θ=1154{\csc \theta = \frac{1}{\frac{\sqrt{15}}{4}}}. Simplifying the expression, you get: csc⁡θ=415{\csc \theta = \frac{4}{\sqrt{15}}}. Rationalizing the denominator, you get: csc⁡θ=41515{\csc \theta = \frac{4\sqrt{15}}{15}}.

Q: How do I find the value of sec⁡θ{\sec \theta} given that cos⁡θ=−14{\cos \theta = -\frac{1}{4}}?

A: To find the value of sec⁡θ{\sec \theta}, you can use the reciprocal identity: sec⁡θ=1cos⁡θ{\sec \theta = \frac{1}{\cos \theta}}. Substituting the value of cos⁡θ{\cos \theta}, you get: sec⁡θ=1−14{\sec \theta = \frac{1}{-\frac{1}{4}}}. Simplifying the expression, you get: sec⁡θ=−4{\sec \theta = -4}.

Q: How do I find the value of cot⁡θ{\cot \theta} given that tan⁡θ=−15{\tan \theta = -\sqrt{15}}?

A: To find the value of cot⁡θ{\cot \theta}, you can use the reciprocal identity: cot⁡θ=1tan⁡θ{\cot \theta = \frac{1}{\tan \theta}}. Substituting the value of tan⁡θ{\tan \theta}, you get: cot⁡θ=1−15{\cot \theta = \frac{1}{-\sqrt{15}}}. Simplifying the expression, you get: cot⁡θ=−115{\cot \theta = -\frac{1}{\sqrt{15}}}. Rationalizing the denominator, you get: cot⁡θ=−1515{\cot \theta = -\frac{\sqrt{15}}{15}}.

Q: What are the relationships between the different trigonometric functions?

A: The relationships between the different trigonometric functions are:

  • Pythagorean Identity: sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1}
  • Reciprocal Identities: csc⁡θ=1sin⁡θ{\csc \theta = \frac{1}{\sin \theta}}, sec⁡θ=1cos⁡θ{\sec \theta = \frac{1}{\cos \theta}}, and cot⁡θ=1tan⁡θ{\cot \theta = \frac{1}{\tan \theta}}
  • Quotient Identity: tan⁡θ=sin⁡θcos⁡θ{\tan \theta = \frac{\sin \theta}{\cos \theta}}

These relationships can be used to find the values of the trigonometric functions given the values of other trigonometric functions.

Q: How do I use the trigonometric functions to solve problems?

A: To use the trigonometric functions to solve problems, you can use the following steps:

  1. Identify the problem: Identify the problem and the information given.
  2. Choose the appropriate trigonometric function: Choose the appropriate trigonometric function to use to solve the problem.
  3. Use the trigonometric identity: Use the trigonometric identity to find the value of the trigonometric function.
  4. Solve the problem: Solve the problem using the value of the trigonometric function.

By following these steps, you can use the trigonometric functions to solve problems involving right triangles, circular functions, and other applications of trigonometry.