3.1.1: Fundamental Trigonometric Identities
- Page ID
- 4171
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Prove equations are true using Reciprocal, Tangent, and other identities.
Basic Trigonometric Identities
The basic trigonometric identities are ones that can be logically deduced from the definitions and graphs of the six trigonometric functions. Previously, some of these identities have been used in a casual way, but now they will be formalized and added to the toolbox of trigonometric identities.
How can you use the trigonometric identities to simplify the following expression?
\(\left[\dfrac{\sin(\dfrac{\pi}{2}−\theta ) }{\sin(−\theta )}\right]^{−1}\)
Trigonometric Identities
An identity is a mathematical sentence involving the symbol “=” that is always true for variables within the domains of the expressions on either side.
Reciprocal Identities
The reciprocal identities refer to the connections between the trigonometric functions like sine and cosecant. Sine is opposite over hypotenuse and cosecant is hypotenuse over opposite. This logic produces the following six identities.
- \(\sin\theta =\dfrac{1}{\csc\theta}\)
- \(\cos\theta =\dfrac{1}{\sec\theta}\)
- \(\tan\theta =\dfrac{1}{\cot\theta}\)
- \(\cot\theta =\dfrac{1}{\tan\theta}\)
- \(\sec\theta =\dfrac{1}{\cos\theta}\)
- \(\csc\theta =\dfrac{1}{\sin\theta}\)
Quotient Identities
The quotient identities follow from the definition of sine, cosine and tangent.
- \(\tan\theta =\sin\theta \cos\theta\)
- \(\cot\theta =\cos\theta \sin\theta\)
Odd/Even Identities
The odd-even identities follow from the fact that only cosine and its reciprocal secant are even and the rest of the trigonometric functions are odd.
- \(\sin(−\theta )=−\sin\theta\)
- \(\cos(−\theta )=\cos\theta\)
- \(\tan(−\theta )=−\tan\theta\)
- \(\cot(−\theta )=−\cot\theta\)
- \(\sec(−\theta )=\sec\theta\)
- \(\csc(−\theta )=−\csc\theta\)
Cofunction Identities
The cofunction identities make the connection between trigonometric functions and their “co” counterparts like sine and cosine. Graphically, all of the cofunctions are reflections and horizontal shifts of each other.
- \(\cos\left(\dfrac{\pi}{2}−\theta \right)=\sin\theta\)
- \(\sin\left(\dfrac{\pi}{2}−\theta \right)=\cos\theta\)
- \(\tan\left(\dfrac{\pi}{2}−\theta \right)=\cot\theta\)
- \(\cot\left(\dfrac{\pi}{2}−\theta \right)=\tan\theta\)
- \(\sec\left(\dfrac{\pi}{2}−\theta \right)=\csc\theta\)
- \(\csc\left(\dfrac{\pi}{2}−\theta \right)=\sec\theta\)
Earlier, you were asked how you could simplify the trigonometric expression:
\(\left[ \dfrac{\sin\left(\dfrac{\pi}{2}−\theta \right)}{\sin(−\theta )}\right]^{−1}\)
Solution
It can be simplified to be equivalent to negative tangent as shown below:
\(\begin{aligned} \left[ \dfrac{\sin(\dfrac{\pi}{2}−\theta )}{\sin(−\theta )} \right]^{−1}&=\dfrac{\sin(−\theta )}{\sin(\dfrac{\pi}{2}−\theta )} \\&=−\sin\theta \cos\theta \\ &=−\tan\theta\end{aligned}\)
If \(\sin\theta =0.87\), find \(\cos \left(\theta −\dfrac{\pi}{2}\right)\).
Solution
While it is possible to use a calculator to find \theta , using identities works very well too.
First you should factor out the negative from the argument. Next you should note that cosine is even and apply the odd-even identity to discard the negative in the argument. Lastly recognize the cofunction identity.
\(\cos\left(\theta −\dfrac{\pi}{2}\right)=\cos\left(−\left(\dfrac{\pi}{2}−\theta \right)\right)=\cos\left(\dfrac{\pi}{2}−\theta \right)=\sin\theta =0.87\)
If \(\cos \left(\theta −\dfrac{\pi}{2}\right)=0.68\) then determine \(\csc(−\theta )\).
Solution
You need to show that \(\cos \left(\theta −\dfrac{\pi}{2}\right)=\cos \left(\dfrac{\pi}{2}−\theta \right)\).
\(\begin{aligned} 0.68&=\cos \left(\theta −\dfrac{\pi}{2}\right)\\&=\cos\left(\dfrac{\pi}{2}−\theta \right) \\&=\sin(\theta ) \end{aligned}\)
Then, \(\csc(−\theta )=−\csc \theta\)
\(\begin{aligned} &=−\dfrac{1}{\sin\theta} \\ &=−(0.68)^{−1} \approx −1.47\end{aligned}\)
Use identities to prove the following: \(\cot(−\beta ) \cot \left(\dfrac{\pi}{2}−\beta \right) \sin(−\beta )= \cos \left(\beta −\dfrac{\pi}{2}\right)\).
Solution
When doing trigonometric proofs, it is vital that you start on one side and only work with that side until you derive what is on the other side. Sometimes it may be helpful to work from both sides and find where the two sides meet, but this work is not considered a proof. You will have to rewrite your steps so they follow from only one side. In this case, work with the left side and keep rewriting it until you have \(\cos \left(\beta −\dfrac{\pi}{2}\right)\).
\(\begin{aligned} \cot(−\beta ) \cot\left(\dfrac{\pi}{2}−\beta \right) \sin(−\beta ) \\ &=−\cot \beta \tan\beta \cdot −\sin\beta \\&=−1\cdot −sin\beta \\&=sin\beta \\&=\cos\left(\dfrac{\pi}{2}−\beta \right) \\&=\cos\left(−\left(\beta −\dfrac{\pi}{2}\right)\right) \\&=\cos\left(\beta −\dfrac{\pi}{2}\right) \end{aligned}\)
Prove the following trigonometric identity by working with only one side.
\(\cos x\sin x\tan x\cot x\sec x\csc x=1\)
Solution
\(\begin{aligned} \cos x\sin x\tan x\cot x\sec x\csc x &=\cos x\sin x\tan x\cdot \dfrac{1}{\tan x} \cdot \dfrac{1}{\cos x}\cdot \dfrac{1}{\sin x} &=1\end{aligned}\)
Review
- Prove the quotient identity for cotangent using sine and cosine.
- Explain why \(\cos\left(\dfrac{\pi}{2}−\theta \right)=\sin\theta\) using graphs and transformations.
- Explain why \(\sec \theta =\dfrac{1}{\cos\theta}\).
- Prove that \(\tan\theta \cdot cot\theta =1\).
- Prove that \(\sin\theta \cdot \csc\theta =1\).
- Prove that \(\sin\theta \cdot sec\theta =\tan\theta \).
- Prove that \(\cos\theta \cdot \csc\theta =\cot\theta \).
- If \(\sin\theta =0.81\), what is \(\sin(−\theta )\)?
- If \(\cos\theta =0.5\), what is \(\cos(−\theta )\)?
- If \(\cos\theta =0.25\), what is \(\sec(−\theta )\)?
- If \(\csc\theta =0.7\), what is \(\sin(−\theta )\)?
- How can you tell from a graph if a function is even or odd?
- Prove \(\tan x\cdot \sec x\csc x\cdot \cot x=\tan x\).
- Prove \(\sin 2x\cdot \sec x\tan x\cdot \csc x=1\).
- Prove \(\cos x\cdot \tan x=\sin x\).
Review (Answers)
To see the Review answers, open this PDF file and look for section 6.1.
Vocabulary
Term | Definition |
---|---|
cofunction | Cofunctions are functions that are identical except for a reflection and horizontal shift. Examples include: sine and cosine, tangent and cotangent, secant and cosecant. |
even | An even function is a function with a graph that is symmetric with respect to the y-axis and has the property that \(f(−x)=f(x)\). |
identity | An identity is a mathematical sentence involving the symbol “=” that is always true for variables within the domains of the expressions on either side. |
Odd Function | An odd function is a function with the property that \(f(−x)=−f(x)\). Odd functions have rotational symmetry about the origin. |
proof | A proof is a series of true statements leading to the acceptance of truth of a more complex statement. |
Additional Resources
Video: Trigonometric Identities - Overview
Practice: Fundamental Trigonometric Identities