3.2.4: Simpler Form of Trigonometric Equations
- Page ID
- 4230
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Breakdown of complex expressions using trigonometric identities.
Sometimes things are simpler than they look. For example, trigonometric identities can sometimes be reduced to simpler forms by applying other rules. For example, can you find a way to simplify \(\dfrac{\cos^3\theta =3\cos\theta +\cos^3\theta}{4}\)?
Trigonometric Equations
By this time in your school career you have probably seen trigonometric functions represented in many ways: ratios between the side lengths of right triangles, as functions of coordinates as one travels along the unit circle and as abstract functions with graphs. Now it is time to make use of the properties of the trigonometric functions to gain knowledge of the connections between the functions themselves. The patterns of these connections can be applied to simplify trigonometric expressions and to solve trigonometric equations.
In order to do this, look for parts of the complex trigonometric expression that might be reduced to fewer trig functions if one of the identities you already know is applied to the expression. As you apply identities, some complex trig expressions have parts that can be cancelled out, others can be reduced to fewer trig functions. Observe how this is accomplished in the examples below.
Simplifying Expressions
1. Simplify the following expression using the basic trigonometric identities: \(\dfrac{1+\tan^2 x}{\csc 2x}\)
\(\begin{aligned}
\dfrac{1+\tan ^{2} x}{\csc ^{2} x} & \ldots\left(1+\tan ^{2} x=\sec ^{2} x\right) \text { Pythagorean Identity } \\
\dfrac{\sec ^{2} x}{\csc ^{2} x} & \ldots\left(\sec ^{2} x=\dfrac{1}{\cos ^{2} x} \text { and } \csc ^{2} x=\dfrac{1}{\sin ^{2} x}\right) \text { Reciprocal Identity } \\ \dfrac{\dfrac{1}{\cos ^{2} x}}{\dfrac{1}{\sin ^{2} x}}&=\left(\dfrac{1}{\cos ^{2} x}\right) \div \left(\dfrac{1}{\sin ^{2} x}\right)\\
\left(\dfrac{1}{\cos ^{2} x}\right) \cdot \left(\dfrac{1}{\sin ^{2} x}\right) &=\dfrac{\sin^2 x}{\cos^2 x}\\
&=\tan ^{2} x \rightarrow \text { Quotient Identity }
\end{aligned}\)
2. Simplify the following expression using the basic trigonometric identities: \(\dfrac{\sin ^2 x+\tan^2 x+\cos ^2 x }{\sec x}\)
\(\begin{aligned}
\dfrac{\sin ^{2} x+\tan ^{2} x+\cos ^{2} x}{\sec x} & \ldots\left(\sin ^{2} x+\cos ^{2} x=1\right) \text { Pythagorean Identity } \\
\dfrac{1+\tan ^{2} x}{\sec x} & \ldots\left(1+\tan ^{2} x=\sec ^{2} x\right) \text { Pythagorean Identity } \\
\dfrac{\sec ^{2} x}{\sec x} &=\sec x
\end{aligned}\)
3. Simplify the following expression using the basic trigonometric identities: \(\cos x−\cos ^3 x\)
\(\begin{aligned}
&\cos x-\cos ^{3} x\\
&\cos x\left(1-\cos ^{2} x\right) \quad \ldots \text { Factor out } \cos x \text { and } \sin ^{2} x=1-\cos ^{2} x\\
&\cos x\left(\sin ^{2} x\right)
\end{aligned}\)
Earlier, you were asked to simplify \(\cos^3\theta =\dfrac{3 \cos\theta +\cos^3\theta}{4}\).
Solution
The easiest way to start is to recognize the triple angle identity:
\(\cos^3\theta =\cos^3\theta −3\sin^2\theta \cos\theta\)
Substituting this into the original equation gives:
\(\cos^3\theta =\dfrac{3\cos\theta +(\cos^3\theta −3\sin^2\theta cos\theta )}{4}\)
Notice that you can then multiply by four and subtract a \(\cos^3\theta\) term:
\(3\cos^3\theta =3\cos\theta −3\sin^2\theta cos\theta\)
And finally pulling out a three and dividing:
\(\cos^3\theta =\cos\theta −\sin^2\theta cos\theta\)
Then pulling out a \(\cos \theta\) and dividing:
\(\cos^2\theta =1−\sin^2\theta\)
Simplify \(\tan^3(x)\csc^3(x)\).
Solution
\(\begin{aligned} \tan^3(x) \csc^3(x)&\\&=\sin^3(x)\cos^3(x)\times \dfrac{ 1}{\sin^3(x)} \\ &=\dfrac{1}{\cos^3(x)} \\ &=sec^3(x) \end{aligned}\)
Show that \(\cot^2(x)+1=\csc^2(x)\).
Solution
Start with \(\sin^2(x)+\cos^2(x)=1\), and divide everything through by \(sin^2(x)\):
\(\begin{aligned} \sin^2(x)+\cos^2(x)=1 \\ &=\dfrac{\sin^2(x)}{\sin^2(x)}+\dfrac{\cos^2(x)}{\sin^2(x)}\\&=\dfrac{1}{\sin^2(x)} \\&=1+\cot^2(x)=\csc^2(x)\end{aligned}\)
Simplify \(\dfrac{\csc^2(x)−1}{\csc^2(x)}\).
Solution
\(\dfrac{\csc^2(x)−1}{\csc^2(x)}\)
Using \(\cot^2(x)+1=\csc^2(x)\) that was proven in #2, you can find the relationship: \(cot^2(x)=\csc^2(x)−1\), you can substitute into the above expression to get:
\(\begin{aligned} \dfrac{\cot^2(x)}{\csc^2(x)} &=\dfrac{\dfrac{\cos^2(x)}{\sin^2(x)}}{\dfrac{1}{\sin^2(x)}} \\&=\cos^2(x)\end{aligned}\)
Review
Simplify each trigonometric expression as much as possible.
- \(\sin(x) \cot(x)\)
- \(\cos(x) tan(x)\)
- \(\dfrac{1+\tan(x)}{1+\cot(x)}\)
- \(\dfrac{1−\sin^2(x)}{1+\sin(x)}\)
- \(\dfrac{\sin^2(x)}{1+\cos(x)}\)
- \((1+\tan^2(x))(\sec^2(x))\)
- \(\sin(x)(\tan(x)+\cot(x))\)
- \(\dfrac{\sec(x)}{\sin(x)}−\dfrac{\sin(x)}{\cos(x)}\)
- \(\dfrac{\sin(x)}{\cot^2(x)}−\dfrac{\sin(x)}{\cos^2(x)}\)
- \(\dfrac{1+\sin(x)}{cos(x)}−\sec(x)\)
- \(\dfrac{\sin^2(x)−\sin^4(x)}{\cos^2(x)}\)
- \(\dfrac{\tan(x)}{\csc^2(x)}+\dfrac{\tan(x)}{\sec^2(x)}\)
- \(\sqrt{1−\cos^2(x)}\)
- \((1−\sin^2(x))(\cos(x))\)
- \((\sec^2(x)+\csc^2(x))−(\tan^2(x)+\cot^2(x))\)
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.3.
Vocabulary
Term | Definition |
---|---|
Trigonometric Identity | A trigonometric identity is an equation that relates two or more trigonometric functions. |
Additional Resources
Video: Example: Solving a Trigonometric Equation Using a Trig Substitution and Factoring
Practice: Simpler Form of Trigonometric Equations