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The Other Trigonometric Functions

Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent of

Finding Exact Values of the Trigonometric Functions Secant, Cosecant, Tangent, and Cotangent

To define the remaining functions, we will once again draw a unit circle with a point \((x,y)\) corresponding to an angle of \(t,\) As with the sine and cosine, we can use the \((x,y)\) coordinates to find the other functions.

The first function we will define is the tangent. The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. In , the tangent of angle \(t\) is equal to \(\frac{y}{x},x\ne 0.\) Because the y-value is equal to the sine of \(\ t,\) and the x-value is equal to the cosine of \(\ t,\) the tangent of angle \(t\) can also be defined as \(\frac{\sin \ t}{\cos \ t},\cos \ t\ne 0.\) The tangent function is abbreviated as \(\text{tan}\text{.}\) The remaining three functions can all be expressed as reciprocals of functions we have already defined.

  • The secant function is the reciprocal of the cosine function. In , the secant of angle \(t\) is equal to \(\frac{1}{\cos \ t}=\frac{1}{x},x\ne 0.\) The secant function is abbreviated as \(\text{sec}\text{.}\)
  • The cotangent function is the reciprocal of the tangent function. In , the cotangent of angle \(t\) is equal to \(\frac{\cos \ t}{\sin \ t}=\frac{x}{y},\ y\ne 0.\) The cotangent function is abbreviated as \(\text{cot}\text{.}\)
  • The cosecant function is the reciprocal of the sine function. In , the cosecant of angle \(t\) is equal to \(\frac{1}{\sin \ t}=\frac{1}{y},y\ne 0.\) The cosecant function is abbreviated as \(\text{csc}\text{.}\)
Example

Try it.

The point \((-\frac{\sqrt{3}}{2},\frac{1}{2})\) is on the unit circle, Find \(\sin \ t,\cos \ t,\tan \ t,\text{sec}\ t,\text{csc}\ t,\) and \(\text{cot}\ t.\)

Solution

Because we know the \((x,y)\) coordinates of the point on the unit circle indicated by angle \(t,\) we can use those coordinates to find the six functions:

\[\begin{array}{l}\sin \ t=y=\frac{1}{2} \\ \cos \ t=x=-\frac{\sqrt{3}}{2} \\ \tan \ t=\frac{y}{x}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{2}(-\frac{2}{\sqrt{3}})=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3} \\ \text{sec}\ t=\frac{1}{x}=\frac{1}{\frac{-\frac{\sqrt{3}}{2}}{}}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3} \\ \text{csc}\ t=\frac{1}{y}=\frac{1}{\frac{1}{2}}=2 \\ \text{cot}\ t=\frac{x}{y}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\frac{\sqrt{3}}{2}(\frac{2}{1})=-\sqrt{3}\end{array}\]

Condensed: the full section is in OpenStax Precalculus 2e.

Using Reference Angles to Evaluate Tangent, Secant, Cosecant, and Cotangent

We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x- and y-values in the original quadrant. shows which functions are positive in which quadrant.

To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “A,” all of the six trigonometric functions are positive. In quadrant II, “Smart,” only sine and its reciprocal function, cosecant, are positive. In quadrant III, “Trig,” only tangent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “Class,” only cosine and its reciprocal function, secant, are positive.

Example

Try it.

Use reference angles to find all six trigonometric functions of \(-\frac{5\pi }{6}.\)

Solution

The angle between this angle’s terminal side and the x-axis is \(\frac{\pi }{6},\) so that is the reference angle. Since \(-\frac{5\pi }{6}\) is in the third quadrant, where both \(x\) and \(y\) are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.

\[\begin{array}{l}\cos (-\frac{5\pi }{6})=-\frac{\sqrt{3}}{2},\sin (-\frac{5\pi }{6})=-\frac{1}{2},\text{tan}(-\frac{5\pi }{6})=\frac{\sqrt{3}}{3} \\ \text{sec}(-\frac{5\pi }{6})=-\frac{2\sqrt{3}}{3},\text{csc}(-\frac{5\pi }{6})=-2,\text{cot}(-\frac{5\pi }{6})=\sqrt{3}\end{array}\]

Condensed: the full section is in OpenStax Precalculus 2e.

Using Even and Odd Trigonometric Functions

To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard.

Consider the function \(f(x)={x}^{2},\) The graph of the function is symmetrical about the y-axis. All along the curve, any two points with opposite x-values have the same function value. This matches the result of calculation: \({(4)}^{2}={(-4)}^{2},\) \({(-5)}^{2}={(5)}^{2},\) and so on. So \(f(x)={x}^{2}\) is an even function, a function such that two inputs that are opposites have the same output. That means \(f(-x)=f(x).\)

Now consider the function \(f(x)={x}^{3},\) The graph is not symmetrical about the y-axis. All along the graph, any two points with opposite x-values also have opposite y-values. So \(f(x)={x}^{3}\) is an odd function, one such that two inputs that are opposites have outputs that are also opposites. That means \(f(-x)=-f(x).\)

We can test whether a trigonometric function is even or odd by drawing a unit circle with a positive and a negative angle, as in . The sine of the positive angle is \(y.\) The sine of the negative angle is −y. The sine function, then, is an odd function. We can test each of the six trigonometric functions in this fashion. The results are

\(\begin{array}{l}\ \sin \ t=y \\ \sin (-t)=-y \\ \sin \ t\ne \sin (-t)\end{array}\) \(\begin{array}{l}\ \text{cos}\ t=x \\ \cos (-t)=x \\ \cos \ t=\cos (-t)\end{array}\) \(\begin{array}{l}\ \text{tan}(t)=\frac{y}{x} \\ \tan (-t)=-\frac{y}{x} \\ \tan \ t\ne \tan (-t)\end{array}\)
\(\begin{array}{l}\ \text{sec}\ t=\frac{1}{x} \\ \text{sec}(-t)=\frac{1}{x} \\ \text{sec}\ t=\text{sec}(-t)\end{array}\) \(\begin{array}{l}\ \text{csc}\ t=\frac{1}{y} \\ \text{csc}(-t)=\frac{1}{-y} \\ \text{csc}\ t\ne \text{csc}(-t)\end{array}\) \(\begin{array}{l}\ \text{cot}\ t=\frac{x}{y} \\ \text{cot}(-t)=\frac{x}{-y} \\ \text{cot}\ t\ne cot(-t)\end{array}\)

Condensed: the full section is in OpenStax Precalculus 2e.

Recognizing and Using Fundamental Identities

We have explored a number of properties of trigonometric functions. Now, we can take the relationships a step further, and derive some fundamental identities. Identities are statements that are true for all values of the input on which they are defined. Usually, identities can be derived from definitions and relationships we already know. For example, the Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine.

Example

Try it.

  1. ⓐ Given \(\sin (45^{\circ})=\frac{\sqrt{2}}{2},\cos (45^{\circ})=\frac{\sqrt{2}}{2},\) evaluate \(\tan (45^{\circ}).\)
  2. ⓑ Given \(\sin (\frac{5\pi }{6})=\frac{1}{2},\text{cos}(\frac{5\pi }{6})=-\frac{\sqrt{3}}{2},\text{evaluate}\ \text{sec}(\frac{5\pi }{6}).\)
Solution

Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions.

ⓐ
\[\begin{array}{l}\tan (45^{\circ})=\frac{\sin (45^{\circ})}{\cos (45^{\circ})} \\ =\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \\ =1\end{array}\]

ⓑ
\[\begin{array}{l}\text{sec}(\frac{5\pi }{6})=\frac{1}{\cos (\frac{5\pi }{6})} \\ =\frac{1}{-\frac{\sqrt{3}}{2}} \\ \begin{array}{l}\ =\frac{-2}{\sqrt{3}} \\ =-\frac{2\sqrt{3}}{3}\end{array}\end{array}\]

Example

Try it.

Simplify \(\frac{\text{sec}\ t}{\tan \ t}.\)

Solution

We can simplify this by rewriting both functions in terms of sine and cosine.

\[\begin{array}{ll}\frac{\text{sec}\ t}{\tan \ t}=\frac{\frac{1}{\cos \ t}}{\frac{\sin \ t}{\cos \ t}} & \text{To divide the functions, we multiply by the reciprocal}. \\ =\frac{1}{\cos \ t}\frac{\cos \ t}{\sin \ t} & \text{Divide out the cosines}. \\ =\frac{1}{\sin \ t} & \text{Simplify and use the identity}. \\ =\text{csc}\ t & \end{array}\]

By showing that \(\frac{\text{sec}\ t}{\tan \ t}\) can be simplified to \(\text{csc}\ t,\) we have, in fact, established a new identity.

\[\frac{\text{sec}\ t}{\tan \ t}=\text{csc}\ t\]

Condensed: the full section is in OpenStax Precalculus 2e.

Evaluating Trigonometric Functions with a Calculator

We have learned how to evaluate the six trigonometric functions for the common first-quadrant angles and to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. If the calculator has a degree mode and a radian mode, confirm the correct mode is chosen before making a calculation.

Evaluating a tangent function with a scientific calculator as opposed to a graphing calculator or computer algebra system is like evaluating a sine or cosine: Enter the value and press the TAN key. For the reciprocal functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must be evaluated as the reciprocal of a sine, cosine, or tangent.

If we need to work with degrees and our calculator or software does not have a degree mode, we can enter the degrees multiplied by the conversion factor \(\frac{\pi }{180}\) to convert the degrees to radians. To find the secant of \(30^{\circ},\) we could press

\[\text{(for a scientific calculator):}\frac{1}{30\times \frac{\pi }{180}}\text{COS}\]

or

\[\text{(for a graphing calculator):}\ \frac{1}{\cos (\frac{30\pi }{180})}\]
Example

Try it.

Evaluate the cosecant of \(\frac{5\pi }{7}.\)

Solution

For a scientific calculator, enter information as follows:

\[\text{1 / ( 5 }\times \ \pi \text{ / 7 ) SIN }=\]\[\text{csc}(\frac{5\pi }{7})\approx 1.279\]

Key Equations

Tangent function \(\tan \ t=\frac{\sin t}{\cos t}\)
Secant function \(\text{sec}\ t=\frac{1}{\cos t}\)
Cosecant function \(\text{csc}\ t=\frac{1}{\sin t}\)
Cotangent function \(\text{cot}\ t=\frac{1}{\tan \ t}=\frac{\cos \ t}{\sin \ t}\)

Key Concepts

  • The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle.
  • The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the reciprocal of the sine function.
  • The six trigonometric functions can be found from a point on the unit circle.
  • Trigonometric functions can also be found from an angle.
  • Trigonometric functions of angles outside the first quadrant can be determined using reference angles.
  • A function is said to be even if \(f(-x)=f(x)\) and odd if \(f(-x)=-f(x).\)
  • Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd.
  • Even and odd properties can be used to evaluate trigonometric functions.
  • The Pythagorean Identity makes it possible to find a cosine from a sine or a sine from a cosine.
  • Identities can be used to evaluate trigonometric functions.
  • Fundamental identities such as the Pythagorean Identity can be manipulated algebraically to produce new identities.
  • The trigonometric functions repeat at regular intervals.
  • The period \(P\) of a repeating function \(f\) is the smallest interval such that \(f(x+P)=f(x)\) for any value of \(x.\)
  • The values of trigonometric functions of special angles can be found by mathematical analysis.
  • To evaluate trigonometric functions of other angles, we can use a calculator or computer software.

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  1. The point \((-\frac{\sqrt{3}}{2},\frac{1}{2})\) is on the unit circle, Find \(\sin \ t,\cos \ t,\tan \ t,\text{sec}\ t,\text{csc}\ t,\) and \(\text{cot}\ t.\)

    Tampilkan jawaban

    Because we know the \((x,y)\) coordinates of the point on the unit circle indicated by angle \(t,\) we can use those coordinates to find the six functions:

    \[\begin{array}{l}\sin \ t=y=\frac{1}{2} \\ \cos \ t=x=-\frac{\sqrt{3}}{2} \\ \tan \ t=\frac{y}{x}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{2}(-\frac{2}{\sqrt{3}})=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3} \\ \text{sec}\ t=\frac{1}{x}=\frac{1}{\frac{-\frac{\sqrt{3}}{2}}{}}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3} \\ \text{csc}\ t=\frac{1}{y}=\frac{1}{\frac{1}{2}}=2 \\ \text{cot}\ t=\frac{x}{y}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\frac{\sqrt{3}}{2}(\frac{2}{1})=-\sqrt{3}\end{array}\]

  2. The point \((\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\) is on the unit circle, Find \(\sin \ t,\cos \ t,\tan \ t,\text{sec}\ t,\text{csc}\ t,\) and \(\text{cot}\ t.\)

    Tampilkan jawaban

    \(\sin \ t=-\frac{\sqrt{2}}{2}\) , \(\cos \ t=\frac{\sqrt{2}}{2}\), \(\tan \ t=-1\), \(\text{sec}\ t=\sqrt{2}\), \(\text{csc}\ t=-\sqrt{2}\), \(\text{cot}\ t=-1\)

  3. Find \(\sin \ t,\cos \ t,\tan \ t,\text{sec}\ t,\text{csc}\ t,\) and \(\text{cot}\ t\) when \(t=\frac{\pi }{6}.\)

    Tampilkan jawaban

    We have previously used the properties of equilateral triangles to demonstrate that \(\sin \ \frac{\pi }{6}=\frac{1}{2}\) and \(\cos \ \frac{\pi }{6}=\frac{\sqrt{3}}{2}.\) We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.

    \[\begin{array}{l}\begin{array}{l} \\ \\ \tan \ \frac{\pi }{6}=\frac{\sin \ \frac{\pi }{6}}{\cos \ \frac{\pi }{6}}\end{array} \\ =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\end{array}\]\[\begin{array}{l}\text{sec}\frac{\pi }{6}=\frac{1}{\cos \frac{\pi }{6}} \\ =\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\end{array}\]\[\text{csc}\frac{\pi }{6}=\frac{1}{\sin \frac{\pi }{6}}=\frac{1}{\frac{1}{2}}=2\]\[\begin{array}{l}\text{cot}\frac{\pi }{6}=\frac{\cos \frac{\pi }{6}}{\sin \frac{\pi }{6}} \\ =\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\end{array}\]

  4. Find \(\sin \ t,\cos \ t,\tan \ t,\text{sec}\ t,\text{csc}\ t,\) and \(\text{cot}\ t\) when \(t=\frac{\pi }{3}.\)

    Tampilkan jawaban

    \(\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\), \(\cos \frac{\pi }{3}=\frac{1}{2}\), \(\tan \frac{\pi }{3}=\sqrt{3}\), \(\text{sec}\frac{\pi }{3}=2\), \(\text{csc}\frac{\pi }{3}=\frac{2\sqrt{3}}{3}\), \(\text{cot}\frac{\pi }{3}=\frac{\sqrt{3}}{3}\)

  5. Use reference angles to find all six trigonometric functions of \(-\frac{5\pi }{6}.\)

    Tampilkan jawaban

    The angle between this angle’s terminal side and the x-axis is \(\frac{\pi }{6},\) so that is the reference angle. Since \(-\frac{5\pi }{6}\) is in the third quadrant, where both \(x\) and \(y\) are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.

    \[\begin{array}{l}\cos (-\frac{5\pi }{6})=-\frac{\sqrt{3}}{2},\sin (-\frac{5\pi }{6})=-\frac{1}{2},\text{tan}(-\frac{5\pi }{6})=\frac{\sqrt{3}}{3} \\ \text{sec}(-\frac{5\pi }{6})=-\frac{2\sqrt{3}}{3},\text{csc}(-\frac{5\pi }{6})=-2,\text{cot}(-\frac{5\pi }{6})=\sqrt{3}\end{array}\]

  6. Use reference angles to find all six trigonometric functions of \(-\frac{7\pi }{4}.\)

    Tampilkan jawaban

    \(\sin (\frac{-7\pi }{4})=\frac{\sqrt{2}}{2},\cos (\frac{-7\pi }{4})=\frac{\sqrt{2}}{2},\tan (\frac{-7\pi }{4})=1,\)
    \(\text{sec}(\frac{-7\pi }{4})=\sqrt{2},\text{csc}(\frac{-7\pi }{4})=\sqrt{2},\text{cot}(\frac{-7\pi }{4})=1\)

  7. If the secant of angle \(t\) is 2, what is the secant of \(-t?\)

    Tampilkan jawaban

    Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle t is 2, the secant of \(-t\) is also 2.

  8. If the cotangent of angle \(t\) is \(\sqrt{3},\) what is the cotangent of \(-t?\)

    Tampilkan jawaban

    \(-\sqrt{3}\)

    1. ⓐ Given \(\sin (45^{\circ})=\frac{\sqrt{2}}{2},\cos (45^{\circ})=\frac{\sqrt{2}}{2},\) evaluate \(\tan (45^{\circ}).\)
    2. ⓑ Given \(\sin (\frac{5\pi }{6})=\frac{1}{2},\text{cos}(\frac{5\pi }{6})=-\frac{\sqrt{3}}{2},\text{evaluate}\ \text{sec}(\frac{5\pi }{6}).\)
    Tampilkan jawaban

    Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions.

    ⓐ
    \[\begin{array}{l}\tan (45^{\circ})=\frac{\sin (45^{\circ})}{\cos (45^{\circ})} \\ =\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \\ =1\end{array}\]

    ⓑ
    \[\begin{array}{l}\text{sec}(\frac{5\pi }{6})=\frac{1}{\cos (\frac{5\pi }{6})} \\ =\frac{1}{-\frac{\sqrt{3}}{2}} \\ \begin{array}{l}\ =\frac{-2}{\sqrt{3}} \\ =-\frac{2\sqrt{3}}{3}\end{array}\end{array}\]

  10. Evaluate \(\text{csc}(\frac{7\pi }{6}).\)

    Tampilkan jawaban

    \(-2\)

  11. Simplify \(\frac{\text{sec}\ t}{\tan \ t}.\)

    Tampilkan jawaban

    We can simplify this by rewriting both functions in terms of sine and cosine.

    \[\begin{array}{ll}\frac{\text{sec}\ t}{\tan \ t}=\frac{\frac{1}{\cos \ t}}{\frac{\sin \ t}{\cos \ t}} & \text{To divide the functions, we multiply by the reciprocal}. \\ =\frac{1}{\cos \ t}\frac{\cos \ t}{\sin \ t} & \text{Divide out the cosines}. \\ =\frac{1}{\sin \ t} & \text{Simplify and use the identity}. \\ =\text{csc}\ t & \end{array}\]

    By showing that \(\frac{\text{sec}\ t}{\tan \ t}\) can be simplified to \(\text{csc}\ t,\) we have, in fact, established a new identity.

    \[\frac{\text{sec}\ t}{\tan \ t}=\text{csc}\ t\]

  12. Simplify \((\tan \ t)(\cos \ t).\)

    Tampilkan jawaban

    \(\sin \ t\)

  13. If \(\text{cos}(t)=\frac{12}{13}\) and \(t\) is in quadrant IV, as shown in , find the values of the other five trigonometric functions.

    Tampilkan jawaban

    We can find the sine using the Pythagorean Identity, \({\cos }^{2}t+{\sin }^{2}t=1,\) and the remaining functions by relating them to sine and cosine.

    \[\begin{array}{l}{(\frac{12}{13})}^{2}+{\sin }^{2}t=1 \\ {\sin }^{2}t=1-{(\frac{12}{13})}^{2} \\ {\sin }^{2}t=1-\frac{144}{169} \\ {\sin }^{2}t=\frac{25}{169} \\ \sin \ t=\pm \sqrt{\frac{25}{169}} \\ \sin \ t=\pm \frac{\sqrt{25}}{\sqrt{169}} \\ \sin \ t=\pm \frac{5}{13}\end{array}\]

    The sign of the sine depends on the y-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y-values are negative, its sine is negative, \(-\frac{5}{13}.\)

    The remaining functions can be calculated using identities relating them to sine and cosine.

    \[\begin{array}{l}\begin{array}{l}\begin{array}{l}\tan \ t=\frac{\sin \ t}{\cos \ t}=\frac{-\frac{5}{13}}{\frac{12}{13}}=-\frac{5}{12} \\ \begin{array}{l} \\ \text{sec}\ t=\frac{1}{\cos \ t}=\frac{1}{\frac{12}{13}}=\frac{13}{12}\end{array}\end{array} \\ \text{csc}\ t=\frac{1}{\sin \ t}=\frac{1}{-\frac{5}{13}}=-\frac{13}{5}\end{array} \\ \text{cot}\ t=\frac{1}{\tan \ t}=\frac{1}{-\frac{5}{12}}=-\frac{12}{5}\end{array}\]

  14. If \(\text{sec}(t)=-\frac{17}{8}\) and 0

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    \(\cos \ t=-\frac{8}{17},\sin \ t=\frac{15}{17},\tan \ t=-\frac{15}{8}\)
    \(\text{csc}\ t=\frac{17}{15},\text{cot}\ t=-\frac{8}{15}\)

  15. Find the values of the six trigonometric functions of angle \(t\) based on .

    Tampilkan jawaban
    \[\begin{array}{l}\sin \ t=y=-\frac{\sqrt{3}}{2} \\ \cos \ t=x=-\frac{1}{2} \\ \tan \ t=\frac{\sin t}{\cos t}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3} \\ \text{sec}\ t=\frac{1}{\cos t}=\frac{1}{-\frac{1}{2}}=-2 \\ \text{csc}\ t=\frac{1}{\sin t}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2\sqrt{3}}{3} \\ \text{cot}\ t=\frac{1}{\tan t}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\end{array}\]

  16. Find the values of the six trigonometric functions of angle \(t\) based on .

    Tampilkan jawaban

    \(\begin{array}{l}\sin \ t=-1,\cos \ t=0,\tan \ t=\text{Undefined} \\ \text{sec}\ t=\text{\,Undefined,}\text{csc}\ t=-1,\text{cot}\ t=0\end{array}\)

  17. If \(\sin (t)=-\frac{\sqrt{3}}{2}\) and \(\text{cos}(t)=\frac{1}{2},\) find \(\text{sec}(t),\text{csc}(t),\text{tan}(t),\text{ cot}(t).\)

    Tampilkan jawaban
    \[\begin{array}{l}\begin{array}{l} \\ \text{sec}\ t=\frac{1}{\cos \ t}=\frac{1}{\frac{1}{2}}=2\end{array} \\ \text{csc}\ t=\frac{1}{\sin \ t}=\frac{1}{-\frac{\sqrt{3}}{2}}-\frac{2\sqrt{3}}{3} \\ \tan \ t=\frac{\sin \ t}{\cos \ t}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3} \\ \text{cot}\ t=\frac{1}{\tan \ t}=\frac{1}{-\sqrt{3}}=-\frac{\sqrt{3}}{3}\end{array}\]

  18. If \(\sin (t)=\frac{\sqrt{2}}{2}\) and \(\cos (t)=\frac{\sqrt{2}}{2},\) find \(\text{sec}(t),\text{csc}(t),\text{tan}(t),\text{ and cot}(t).\)

    Tampilkan jawaban

    \(\text{sec}\ t=\sqrt{2},\text{csc}\ t=\sqrt{2},\tan \ t=1,\text{cot}\ t=1\)

  19. Evaluate the cosecant of \(\frac{5\pi }{7}.\)

    Tampilkan jawaban

    For a scientific calculator, enter information as follows:

    \[\text{1 / ( 5 }\times \ \pi \text{ / 7 ) SIN }=\]\[\text{csc}(\frac{5\pi }{7})\approx 1.279\]

  20. Evaluate the cotangent of \(-\frac{\pi }{8}.\)

    Tampilkan jawaban

    \(\approx -2.414\)

  21. On an interval of \([0,2\pi ),\) can the sine and cosine values of a radian measure ever be equal? If so, where?

    Tampilkan jawaban

    Yes, when the reference angle is \(\frac{\pi }{4}\) and the terminal side of the angle is in quadrants I and III. Thus, at \(x=\frac{\pi }{4},\frac{5\pi }{4},\) the sine and cosine values are equal.

  22. What would you estimate the cosine of \(\pi\) degrees to be? Explain your reasoning.

  23. For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle?

    Tampilkan jawaban

    Substitute the sine of the angle in for \(y\) in the Pythagorean Theorem \({x}^{2}+{y}^{2}=1.\) Solve for \(x\) and take the negative solution.

  24. Describe the secant function.

  25. Tangent and cotangent have a period of \(\pi .\) What does this tell us about the output of these functions?

    Tampilkan jawaban

    The outputs of tangent and cotangent will repeat every \(\pi\) units.

  26. \(\tan \ \frac{\pi }{6}\)

  27. \(\text{sec}\ \frac{\pi }{6}\)

    Tampilkan jawaban

    \(\frac{2\sqrt{3}}{3}\)

  28. \(\text{csc}\ \frac{\pi }{6}\)

  29. \(\text{cot}\ \frac{\pi }{6}\)

    Tampilkan jawaban

    \(\sqrt{3}\)

  30. \(\tan \ \frac{\pi }{4}\)

  31. \(\text{sec}\ \frac{\pi }{4}\)

    Tampilkan jawaban

    \(\sqrt{2}\)

  32. \(\text{csc}\ \frac{\pi }{4}\)

  33. \(\text{cot}\ \frac{\pi }{4}\)

    Tampilkan jawaban

    1

  34. \(\tan \ \frac{\pi }{3}\)

  35. \(\text{sec}\ \frac{\pi }{3}\)

    Tampilkan jawaban

    2

  36. \(\text{csc}\ \frac{\pi }{3}\)

  37. \(\text{cot}\ \frac{\pi }{3}\)

    Tampilkan jawaban

    \(\frac{\sqrt{3}}{3}\)

  38. \(\tan \ \frac{5\pi }{6}\)

  39. \(\text{sec}\ \frac{7\pi }{6}\)

    Tampilkan jawaban

    \(-\frac{2\sqrt{3}}{3}\)

  40. \(\text{csc}\ \frac{11\pi }{6}\)

Simpan pekerjaan Anda sendiri

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Daftar Masuk

Simbol yang digunakan di sini

Sentuh simbol apapun untuk definisi lengkap, gambar, dan apa setiap huruf di dalamnya berarti.

Bagaimana: The Other Trigonometric Functions

  1. Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent of
  2. Use reference angles to evaluate the trigonometric functions secant, cosecant, tangent, and cotangent.
  3. Use properties of even and odd trigonometric functions.
  4. Recognize and use fundamental identities.
  5. Evaluate trigonometric functions with a calculator.
  6. The
  7. The
  8. The

Pertanyaan yang sering diajukan

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Bagian dari halaman ini diadaptasi dari OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Diringkas dan dijelaskan ulang di sini; kesalahan adalah tanggung jawab kami. Karena sumbernya dilisensikan untuk penggunaan non-komersial, pelajaran ini gratis, menjual apa-apa, dan dibagi di bawah lisensi yang sama.

Lainnya di Trigonometry

The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions