3.2.3: Proofs of Trigonometric Identities
- Page ID
- 4227
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Convert to sine/cosine, use basic identities, and simplify sides of the equation.
Verify that \(\dfrac{\sin^2 x}{\tan^2 x}=1−\sin^2 x\).
Verifying Trigonometric Identities
Now that you are comfortable simplifying expressions, we will extend the idea to verifying entire identities. Here are a few helpful hints to verify an identity:
- Change everything into terms of sine and cosine.
- Use the identities when you can.
- Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.
Let's verify the following identities.
- \(\dfrac{\cot^2 x}{\csc x}=\csc x−\sin x\)
Rather than have an equal sign between the two sides of the equation, we will draw a vertical line so that it is easier to see what we do to each side of the equation. Start with changing everything into sine and cosine.
\(\begin{array}{l|l}
\dfrac{\cot ^{2} x}{\csc x} & \csc x-\sin x \\
\dfrac{\dfrac{\cos ^{2} x}{\sin ^{2} x}}{\dfrac{1}{\sin x}} & \dfrac{1}{\sin x} \\
\dfrac{\cos ^{2} x}{\sin x} &
\end{array}\)
Now, it looks like we are at an impasse with the left-hand side. Let’s combine the right-hand side by giving them same denominator.
\(\begin{array}{|c}
\dfrac{1}{\sin x}-\dfrac{\sin ^{2} x}{\sin x} \\
\dfrac{1-\sin ^{2} x}{\sin x} \\
\dfrac{\cos ^{2} x}{\sin x}
\end{array}\)
The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, \(\sin^2 \theta +\cos^2 \theta =1\), and isolated the \(\cos^2 x=1−\sin^2x\).
There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to\(\dfrac{\cos^2 x}{\sin^2 x}\), we could have also substituted the identity \(\cot^2 x=\csc^2 x−1\).
- \(\dfrac{\sin x}{1−\cos x}=\dfrac{1+\cos x}{\sin x}\)
Multiply the left-hand side of the equation by \(\dfrac{1+\cos x}{1+\cos x}\).
\(\begin{aligned}
\dfrac{\sin x}{1-\cos x} &=\dfrac{1+\cos x}{\sin x} \\
\dfrac{1+\cos x}{1+\cos x} \cdot \dfrac{\sin x}{1-\cos x} &=\\
\dfrac{\sin (1+\cos x)}{1-\cos ^{2} x} &=\\
\dfrac{\sin (1+\cos x)}{\sin ^{2} x} &=\\
\dfrac{1+\cos x}{\sin x} &=
\end{aligned}\)
The two sides are the same, so we are done.
- \(\sec(−x)=\sec x\)
Change secant to cosine.
\(\sec(−x)=\dfrac{1}{cos(−x)}\)
From the Negative Angle Identities, we know that \(\cos(−x)=\cos x\).
\(\begin{aligned} &=\dfrac{1}{\cos x} \\&=\sec x\end{aligned}\)
Earlier, you were asked to verify that \(\sin^2 x \tan^2 x=1−\sin^2 x\).
Solution
Start by simplifying the left-hand side of the equation.
\(\sin ^2 x \tan ^2 x=\dfrac{\sin ^2 x}{\dfrac{\sin ^2 x}{\cos ^2 x}}=\cos ^2 x\)
Now simplify the right-hand side of the equation. By manipulating the Trigonometric Identity,
\(\sin ^2 x+\cos ^2 x=1\), we get \(\cos ^2 x=1−\sin ^2 x\).
\(\cos ^2 x=\cos ^2 x\) and the equation is verified.
Verify the following identities.
\(\cos x \sec x=1\)
Solution
Change secant to cosine.
\(\begin{aligned} \cos x \sec x&=\cos x \cdot \dfrac{1}{\cos x} \\&=1\end{aligned}\)
\(2−sec^2 x=1−\tan ^2 x\)
Solution
Use the identity \(1+\tan^2 \theta =\sec^2 \theta\).
\(\begin{aligned} 2−\sec^2 x &=2−(1+\tan ^2 x) \\&=2−1−\tan ^2 x \\&=1−\tan ^2 x\end{aligned}\)
\(\dfrac{\cos(−x)}{1+\sin(−x)}=\sec x+\tan x\)
Solution
Here, start with the Negative Angle Identities and multiply the top and bottom by \(\dfrac{1+\sin x}{1+\sin x}\) to make the denominator a monomial.
\(\begin{aligned}
\dfrac{\cos (-x)}{1+\sin (-x)} &=\dfrac{\cos x}{1-\sin x} \cdot \dfrac{1+\sin x}{1+\sin x} \\
&=\dfrac{\cos x(1+\sin x)}{1-\sin ^{2} x} \\
&=\dfrac{\cos x(1+\sin x)}{\cos ^{2} x} \\
&=\dfrac{1+\sin x}{\cos x} \\
&=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x} \\
&=\sec x+\tan x
\end{aligned}\)
Review
Verify the following identities.
- \(\cot(−x)=−\cot x\)
- \(\csc(−x)=−\csc x\)
- \(\tan x \csc x \cos x=1\)
- \(\sin x+\cos x \cot x=\csc x\)
- \(\csc\left(\dfrac{\pi}{2}−x\right)=\sec x\)
- \(\tan\left(\dfrac{\pi}{2}−x\right)=\tan x\)
- \(\dfrac{\csc x}{\sin x}−\dfrac{\cot x}{\tan x}=1\)
- \(\dfrac{\tan ^2 x}{\tan ^2 x+1}=\sin ^2 x\)
- \((\sin x−\cos x)^2+(\sin x+\cos x)^2=2\)
- \(\sin x−\sin x \cos ^2 x=\sin ^3 x\)
- \(\tan ^2 x+1+\tan x \sec x=\dfrac{1+\sin x}{\cos ^2 x}\)
- \(\cos ^2 x=\csc x \cos x \tan x+\cot x\)
- \(\dfrac{1}{1−\sin x}−\dfrac{1}{1+\sin x}=2\tan x \sec x\)
- \(\csc^4 x−\cot^4 x=\csc ^2 x+\cot^2x\)
- \((\sin x−\tan x)(\cos x−\cot x)=(\sin x−1)(\cos x−1)\)
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.9.