Reciprocal and Quotient Identities
The next set of fundamental identities is the set of reciprocal identities , which, as their name implies, relate trigonometric functions that are reciprocals of each other.
Reciprocal Identities
[latex]\sin \theta =\frac{1}{\csc \theta }[/latex]
[latex]\csc \theta =\frac{1}{\sin \theta }[/latex]
[latex]\cos \theta =\frac{1}{\sec \theta }[/latex]
[latex]\sec \theta =\frac{1}{\cos \theta }[/latex]
[latex]\tan \theta =\frac{1}{\cot \theta }[/latex]
[latex]\cot \theta =\frac{1}{\tan \theta }[/latex]
The final set of identities is the set of quotient identities , which define relationships among certain trigonometric functions and can be very helpful in verifying other identities.
Quotient Identities
[latex]\tan \theta =\frac{\sin \theta }{\cos \theta }[/latex]
[latex]\cot \theta =\frac{\cos \theta }{\sin \theta }[/latex]
The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.
trigonometric identities
The Pythagorean identities are based on the properties of a right triangle.
[latex]\begin{gathered} {\cos}^{2}\theta + {\sin}^{2}\theta=1 \\ 1+{\tan}^{2}\theta={\sec}^{2}\theta \\ 1+{\cot}^{2}\theta={\csc}^{2}\theta\end{gathered}[/latex]
The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.
[latex]\begin{gathered} \cos(-\theta)=\cos(\theta) \\\sin(-\theta)=-\sin(\theta) \\\tan(-\theta)=-\tan(\theta) \\\cot(-\theta)=-\cot(\theta) \\\sec(-\theta)=\sec(\theta) \\\csc(-\theta)=-\csc(\theta) \end{gathered}[/latex]
The reciprocal identities define reciprocals of the trigonometric functions.
[latex]\begin{gathered}\sin\theta=\frac{1}{\csc\theta} \\ \cos\theta=\frac{1}{\sec\theta} \\ \tan\theta=\frac{1}{\cot\theta} \\ \cot\theta=\frac{1}{\tan\theta} \\ \sec\theta=\frac{1}{\cos\theta} \\ \csc\theta=\frac{1}{\sin\theta}\end{gathered}[/latex]
The quotient identities define the relationship among the trigonometric functions.
[latex]\begin{gathered} \tan\theta=\frac{\sin\theta}{\cos\theta} \\ \cot\theta=\frac{\cos\theta}{\sin\theta} \end{gathered}[/latex]
Graph both sides of the identity [latex]\cot \theta =\frac{1}{\tan \theta }[/latex]. In other words, on the graphing calculator, graph [latex]y=\cot \theta[/latex] and [latex]y=\frac{1}{\tan \theta }[/latex].
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Analysis of the Solution
We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.
Verify [latex]\tan \theta \cos \theta =\sin \theta[/latex].
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We will start on the left side, as it is the more complicated side:
[latex]\begin{align}\tan \theta \cos \theta &=\left(\frac{\sin \theta }{\cos \theta }\right)\cos \theta \\ &=\left(\frac{\sin \theta }{\cancel{\cos \theta }}\right)\cancel{\cos \theta } \\ &=\sin \theta \end{align}[/latex]
Analysis of the Solution
This identity was fairly simple to verify, as it only required writing [latex]\tan \theta[/latex] in terms of [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex].
Verify the identity [latex]\csc \theta \cos \theta \tan \theta =1[/latex].
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[latex]\begin{align}\csc \theta \cos \theta \tan \theta &=\left(\frac{1}{\sin \theta }\right)\cos \theta \left(\frac{\sin \theta }{\cos \theta }\right) \\ &=\frac{\cos \theta }{\sin \theta }\left(\frac{\sin \theta }{\cos \theta }\right) \\ &=\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta } \\ &=1\end{align}[/latex]
Even and Odd Identities
Even-Odd Identities
[latex]\begin{gathered}\tan \left(-\theta \right)=-\tan \theta\\ \cot \left(-\theta \right)=-\cot \theta \end{gathered}[/latex]
[latex]\begin{gathered}\sin \left(-\theta \right)=-\sin \theta\\ \csc \left(-\theta \right)=-\csc \theta\end{gathered}[/latex]
[latex]\begin{gathered}\cos \left(-\theta \right)=\cos \theta \\ \sec \left(-\theta \right)=\sec \theta \end{gathered}[/latex]
To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.
Verify the following equivalency using the even-odd identities:
[latex]\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]={\cos }^{2}x[/latex]
Show Solution
Working on the left side of the equation, we have
[latex]\begin{align}\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]&=\left(1+\sin x\right)\left(1-\sin x\right)&& \text{Since sin(-}x\text{)=}-\sin x \\ &=1-{\sin }^{2}x&& \text{Difference of squares} \\ &={\cos }^{2}x&& {\text{cos}}^{2}x=1-{\sin }^{2}x\end{align}[/latex]
Show that [latex]\frac{\cot \theta }{\csc \theta }=\cos \theta[/latex].
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[latex]\begin{align}\frac{\cot \theta }{\csc \theta }&=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }} \\ &=\frac{\cos \theta }{\sin \theta }\cdot \frac{\sin \theta }{1} \\ &=\cos \theta \end{align}[/latex]
Create an identity for the expression [latex]2\tan \theta \sec \theta[/latex] by rewriting strictly in terms of sine.
Show Solution
There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:
[latex]\begin{align}2\tan \theta \sec \theta &=2\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{1}{\cos \theta }\right) \\ &=\frac{2\sin \theta }{{\cos }^{2}\theta } \\ &=\frac{2\sin \theta }{1-{\sin }^{2}\theta }&& \text{Substitute }1-{\sin }^{2}\theta \text{ for }{\cos }^{2}\theta \end{align}[/latex]
Thus,
[latex]2\tan \theta \sec \theta =\frac{2\sin \theta }{1-{\sin }^{2}\theta }[/latex]
Verify the identity:
[latex]\begin{align}\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}=\cos \theta -\sin \theta\end{align}[/latex]
Show Solution
Let’s start with the left side and simplify:
[latex]\begin{align}\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}&=\frac{{\left[\sin \left(-\theta \right)\right]}^{2}-{\left[\cos \left(-\theta \right)\right]}^{2}}{\sin \left(-\theta \right)-\cos \left(-\theta \right)} \\ &=\frac{{\left(-\sin \theta \right)}^{2}-{\left(\cos \theta \right)}^{2}}{-\sin \theta -\cos \theta }&& \sin \left(-x\right)=-\sin x\text{ and }\cos \left(-x\right)=\cos x \\ &=\frac{{\left(\sin \theta \right)}^{2}-{\left(\cos \theta \right)}^{2}}{-\sin \theta -\cos \theta }&& \text{Difference of squares} \\ &=\frac{\left(\sin \theta -\cos \theta \right)\left(\sin \theta +\cos \theta \right)}{-\left(\sin \theta +\cos \theta \right)} \\ &=\frac{\left(\sin \theta -\cos \theta \right)\left(\cancel{\sin \theta +\cos \theta }\right)}{-\left(\cancel{\sin \theta +\cos \theta }\right)} \\ &=\cos \theta -\sin \theta\end{align}[/latex]
Verify the identity [latex]\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }=\frac{\sin \theta +1}{\tan \theta }[/latex].
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[latex]\begin{align}\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }&=\frac{\left(\sin \theta +1\right)\left(\sin \theta -1\right)}{\tan \theta \left(\sin \theta -1\right)}\\ &=\frac{\sin \theta +1}{\tan \theta }\end{align}[/latex]
Verify the identity: [latex]\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)=1[/latex].
Show Solutions
We will work on the left side of the equation.
[latex]\begin{align}\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)&=\left(1-{\cos }^{2}x\right)\left(1+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) \\ &=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) && \text{Find the common denominator}. \\ &=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\right) \\ &=\left({\sin }^{2}x\right)\left(\frac{1}{{\sin }^{2}x}\right) \\ &=1\end{align}[/latex]
