Skip to main content
K12 LibreTexts

3.1.3: Reciprocal Identities

  • Page ID
    4174
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Relationship between sine / cosine / tangent and cosecant / secant / cotangent.

    You are already familiar with the trig identities of sine, cosine, and tangent. As you know, any fraction also has an inverse, which is found by reversing the positions of the numerator and denominator.

    Can you list what the ratios would be for the three trig functions (sine, cosine, and tangent) with the numerators and denominators reversed?

    Reciprocal Identities

    A reciprocal of a fraction ab is the fraction ba. That is, we find the reciprocal of a fraction by interchanging the numerator and the denominator, or flipping the fraction. The six trig functions can be grouped in pairs as reciprocals.

    First, consider the definition of the sine function for angles of rotation: \(\sin \theta =\dfrac{y}{r}\). Now consider the cosecant function: \(\csc \theta=\dfrac{r}{y}\). In the unit circle, these values are \(\sin \theta =\dfrac{y}{1}=y\) and \(\csc \theta=\dfrac{1}{y}\). These two functions, by definition, are reciprocals. Therefore the sine value of an angle is always the reciprocal of the cosecant value, and vice versa. For example, if \(\sin \theta =\dfrac{1}{2}\), then \(\csc \theta=\dfrac{2}{1}=2\).

    Analogously, the cosine function and the secant function are reciprocals, and the tangent and cotangent function are reciprocals:

    \(\begin{aligned}
    \sec \theta &=\frac{1}{\cos \theta} & \text { or } & \cos \theta=\frac{1}{\sec \theta} \\
    \cot \theta &=\frac{1}{\tan \theta} & \text { or } & \tan \theta=\frac{1}{\cot \theta}
    \end{aligned}\)

    Using Reciprocal Identities

    Find the value of the following expressions using a reciprocal identity.

    1. \(\cos \theta=.3\), \(\sec \theta=?\)

    \(\sec \theta=\dfrac{10}{3}\)

    These functions are reciprocals, so if \(\cos \theta=.3\), then \(\sec \theta=1.3\). It is easier to find the reciprocal if we express the values as fractions: \(\cos \theta=.3=\dfrac{3}{10} \Rightarrow \sec \theta=103\).

    2. \(\cot \theta=\dfrac{4}{3}\), \(\tan \theta=?\)

    These functions are reciprocals, and the reciprocal of \(\dfrac{4}{3}\) is \(\dfrac{3}{4}\).

    We can also use the reciprocal relationships to determine the domain and range of functions.

    3. \(\sin \theta =\dfrac{1}{2}\), \(\csc \theta=?\)

    These functions are reciprocals, and the reciprocal of \(\dfrac{1}{2}\) is 2.

    Example \(\PageIndex{1}\)

    Earlier, you were asked to list the ratios for the three trigonometric functions with the numerators and denominators reversed.

    Solution

    Since the three regular trig functions are defined as:

    \(\begin{aligned} \sin&=\dfrac{\text{opposite}}{\text{hypotenuse}} \\ \cos&=\dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \tan&=\dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}\)

    then the three functions - called "reciprocal functions" are:

    \(\begin{aligned}\csc&=\dfrac{\text{hypotenuse}}{\text{opposite}} \\ \sec&=\dfrac{\text{hypotenuse}}{\text{adjacent}}\\ \cot&=\dfrac{\text{adjacent}}{\text{opposite}}\end{aligned}\)

    Example \(\PageIndex{2}\)

    State the reciprocal function of cosecant.

    Solution

    The reciprocal function of cosecant is sine.

    Example \(\PageIndex{3}\)

    Find the value of the expression using a reciprocal identity.

    \(\sec \theta=\dfrac{2}{\pi}\), \(\cos \theta=?\)

    Solution

    These functions are reciprocals, and the reciprocal of \(\dfrac{2}{\pi}\) is \(\dfrac{\pi}{2}\).

    Example \(\PageIndex{4}\)

    Find the value of the expression using a reciprocal identity.

    \(\csc \theta=4\), \(\cos \theta=?\)

    Solution

    These functions are reciprocals, and the reciprocal of 4 is \(\dfrac{1}{4}\).

    Review

    1. State the reciprocal function of secant.
    2. State the reciprocal function of cotangent.
    3. State the reciprocal function of sine.

    Find the value of the expression using a reciprocal identity.

    1. \(\sin \theta =\dfrac{1}{2}\), \(\csc \theta=?\)
    2. \(\cos \theta=\dfrac{−\sqrt{3}}{2}\), \(\sec \theta=?\)
    3. \(\tan \theta=1\), \(\cot \theta=?\)
    4. \(\sec \theta=\sqrt{2}\), \(\cos \theta=?\)
    5. \(\csc \theta=2\), \(\sin \theta =?\)
    6. \(\cot \theta=−1\), \(\tan \theta=?\)
    7. \(\sin \theta =\dfrac{\sqrt{3}}{2}\), \(\csc \theta=?\)
    8. \(\cos \theta=0\), \(\sec \theta=?\)
    9. \(\tan \theta=\text{undefined}\), \(\cot \theta=?\)
    10. \(\csc \theta=\dfrac{2\sqrt{3}}{3}\), \(\sin \theta =?\)
    11. \(\sin \theta =\dfrac{−1}{2}\) and \(\tan \theta=\dfrac{\sqrt{3}}{3}\),\(\cos \theta=?\)
    12. \(\cos \theta=\dfrac{\sqrt{2}}{2}\) and \(\tan \theta=1\), \(\sin \theta =?\)

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 1.21.

    Vocabulary

    Term Definition
    domain The domain of a function is the set of x-values for which the function is defined.
    Range The range of a function is the set of y values for which the function is defined.
    Reciprocal Trig Function A reciprocal trigonometric function is a function that is the reciprocal of a typical trigonometric function. For example, since \(\sin x=\dfrac{\text{opposite}}{\text{hypotenuse}}\), the reciprocal function is \(\csc x=\dfrac{\text{hypotenuse}}{\text{opposite}}\)

    Additional Resources

    Interactive Element

    Video: The Reciprocal, Quotient, and Pythagorean Identities


    This page titled 3.1.3: Reciprocal Identities is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.

    CK-12 Foundation
    LICENSED UNDER
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License
    • Was this article helpful?