Opening the book…
Reading Questions
- What are the five main families of basic trigonometric identities summarized in this section?
- What is the first step recommended in the "How To" guide for proving a trigonometric identity?
- When proving an identity, why is it considered "illegal" to perform an operation on both sides of the equals sign?
- If you are stuck while trying to prove an identity, what are two strategies the text suggests trying?
- What does it mean to "keep an eye on the prize" when proving an identity?
- In the proof of \([1+\sin(x)][1+\sin(-x)] = \cos^2(x)\), which two fundamental identities were used?
- What is the general strategy for proving an identity if you are unsure where to begin?
- Explain why it is generally easier to start with the more complex side of an identity.
Homework
Concept Check
-
We showed that \(\sin \left(\theta\right) = \pm \sqrt{1 - \cos^2 \left(\theta\right)}\) for all \(\theta \). Give an example of an angle \(\theta \) such that \(\sin\left(\theta\right) = -\sqrt{1 - \cos^2\left(\theta\right)} \).
-
We showed that \(\cos\left(\theta\right) = \pm \sqrt{1 - \sin^2\left(\theta\right)} \) for all \(\theta \). Give an example of an angle \(\theta \) such that \(\cos\left(\theta\right) = -\sqrt{1 - \sin^2\left(\theta\right)} \).
-
A student claims that \(\cos(\theta) + \sin(\theta) = 1\) is an identity because \(\cos(0) + \sin(0) = 1 + 0 = 0\). How would you respond to this student?
Proofs
For the following exercises, prove the identity. Assume all quantities are defined.
-
\( \tan\left( \theta \right) + \tan\left( -\theta \right) = 0\)
-
\( \cos^2\left( \theta \right) - \sin^2\left( \theta \right) = 2\cos^2\left( \theta \right) - 1 \)
-
\( \cos^2\left( \theta \right) - \sin^2\left( \theta \right) = 1 - 2\sin^2\left( \theta \right) \)
-
\( \dfrac{\sin^2\left( \theta \right)}{1 - \sin^2\left( \theta \right)} = \tan^2\left( \theta \right) \)
-
\( \sin^4\left( \theta \right) - \cos^4\left( \theta \right) = \sin^2\left( \theta \right) - \cos^2\left( \theta \right) \)
-
\( \sin^2\left( x \right) + \tan^2\left( x \right) + \cos^2\left( x \right) = \sec^2\left( x \right) \)
-
\( \cos^4\left( \theta \right) - \sin^4\left( \theta \right) = 1 - 2\sin^2\left( \theta \right) \)
-
\(\dfrac{\sin(x) + \cos(x)}{\cos(x)} = \tan(x) + 1\)
-
\( \dfrac{\cos^2\left( \theta \right)}{1 + \sin\left( \theta \right)} = 1 - \sin\left( \theta \right) \)
-
\(\dfrac{\sin^2(x)}{1 + \cos(x)} = 1 - \cos(x)\)
-
\(\dfrac{\cos(x)}{1 + \sin(x)} = \sec(x) - \tan(x)\)
-
\(\dfrac{\sin(x) - \cos(x)}{\sin(x) + \cos(x)} = \dfrac{\tan(x) - 1}{1 + \tan(x)}\)
-
\( \dfrac{1 - \tan\left( \theta \right)}{1 + \tan\left( \theta \right)} = \dfrac{\cot\left( \theta \right) - 1}{\cot\left( \theta \right) + 1} \)
-
\(\dfrac{1}{\tan(x) + \cot(x)} = \sin(x) \cdot \cos(x)\)
-
\(\dfrac{\tan(x) + \sec(x) - 1}{\tan(x) - \sec(x) + 1} = \dfrac{1 + \sin(x)}{\cos(x)}\)
-
\(\dfrac{1 + \sin(x)}{\cos(x)} = \sec(x) + \tan(x)\)
-
\(\dfrac{\sin(x) - \cos(x)}{\cos(-x) - \sin(-x)} = \dfrac{-1 - \tan(-x)}{1 + \tan(x)}\)
-
\(\dfrac{\sin(-x)}{1 + \cos(x)} = \dfrac{\cos(-x) - 1}{\sin(x)}\)
-
\(\dfrac{\sec(x) - 1}{\sec(x) + 1} = \dfrac{\sin^2(x)}{(\cos(x) + 1)^2}\)
-
\(\dfrac{\sec(x) - 1 - \tan(-x)}{\tan(x) - \sec(-x) + 1} = \dfrac{1 - \sin(-x)}{\cos(x)}\)
-
\(\dfrac{\cos(-x)}{1 - \sin(-x)} = \sec(-x) + \tan(-x)\)
-
\(\dfrac{1 - \sin(-x)}{\cos(x)} = \sec(-x) - \tan(-x)\)
-
\(\sin(3\pi - 2\theta) = -\sin(2\theta - 3\pi)\)
-
\(\dfrac{\cos(x)}{1 - \sin(x)} = \sec(x) + \tan(x)\)
-
\(\dfrac{1 + \cos\left( \frac{\pi}{2} - x \right)}{\cos(x)} = \csc\left( \frac{\pi}{2} - x \right) + \tan(x)\)
-
\(\dfrac{1 - \cos(x)}{\cos(x)} = \dfrac{\sin^2(x)}{\cos(x)(1 + \cos(x))}\)
-
\(\dfrac{\sin^2(x)}{\cos(x)} = \tan(x) \cdot \sin(x)\)
-
\(\dfrac{\sin(-x) - \cos(x)}{\cos(x)} = \tan(-x) - 1\)
-
\( \dfrac{1 - 2\cos^2\left( \theta \right)}{\sin\left( \theta \right) \cos\left( \theta \right)} = \tan\left( \theta \right) - \cot\left( \theta \right)\)
-
\(\cos \left( -\dfrac{\pi}{4} - 5t \right) = \cos \left( 5t + \dfrac{\pi}{4} \right)\)
-
\(\dfrac{\cos\left( \frac{\pi}{2} - x \right)}{\cos(x) + 1} = \dfrac{1 - \sin\left( \frac{\pi}{2} - x \right)}{\sin(x)}\)
-
\( \dfrac{1 - \tan^2\left( \theta \right)}{1 - \cot^2\left( \theta \right)} = 1 - \sec^2\left( \theta \right) \)
-
\(\dfrac{\cos(-x)}{1 + \sin(-x)} = \sec(x) - \tan(-x)\)
-
\(\dfrac{1 + \cos(x)}{\sin(x)} = \csc(x) + \cot(x)\)
-
\(\dfrac{1 - \sin^2(x)}{\cos(x)} = \cos(x)\)
-
\(\dfrac{1 - \sin(x)}{1 + \sin(x)} = \dfrac{\sec(x) - \tan(x)}{\sec(x) + \tan(x)}\)
-
\(\dfrac{\sec(x) - 1}{\sec(x) + 1} = \dfrac{1 - \cos(x)}{1 + \cos(x)}\)
-
\(\dfrac{1 + \sin(x)}{\cos(x)} = \dfrac{\cos(x)}{1 - \sin(x)}\)
-
\(\tan(-t^{2} + 1) = -\tan(t^{2} - 1)\)
-
\(\dfrac{1 - \cos\left( \frac{\pi}{2} - x \right)}{\cos(x)} = \csc\left( \frac{\pi}{2} - x \right) - \tan(x)\)
-
\(\dfrac{\sin(x) - \sin\left(\dfrac{\pi}{2} - x\right)}{\sin(x) + \cos(x)} = \dfrac{\cot\left(\dfrac{\pi}{2} - x\right) - 1}{1 + \tan(x)}\)
-
\(\dfrac{\cot\left( \frac{\pi}{2} - B \right) + \sec(B) - 1}{\tan(B) - \csc\left( \frac{\pi}{2} - B \right) + 1} = \dfrac{1 - \sin(-B)}{\cos(B)}\)
-
\(\dfrac{\sin(\phi)}{1 + \cos(\phi)} = \dfrac{1 - \cos(\phi)}{\sin(\phi)}\)
-
\(\dfrac{\cos^2(\theta)}{1 - \sin(\theta)} = 1 + \sin(\theta)\)
-
\(\csc(-\theta - 5) = -\csc(\theta + 5)\)
-
\(\dfrac{\sin\left(\dfrac{\pi}{2} - \alpha\right)}{1 + \sin(\alpha)} = \csc\left(\dfrac{\pi}{2} - \alpha\right) - \tan(\alpha)\)
-
\(\dfrac{1 + \sin(-\theta)}{1 + \sin(\theta)} = \dfrac{\sec(-\theta) + \tan(-\theta)}{\sec(-\theta) + \tan(\theta)}\)
-
\(\dfrac{1 + \cos(-A)}{\sin(-A)} = -\csc(A) + \cot(-A)\)
-
\(\dfrac{\cos(-\phi) - 1}{\cos(\phi)} = \dfrac{\sin^2(-\phi)}{\cos(\phi)(1 + \cos(\phi))}\)
-
\(\dfrac{\sin\left( \frac{\pi}{2} - \theta \right)}{1 - \sin(\theta)} = \csc\left( \frac{\pi}{2} - \theta \right) + \cot\left( \frac{\pi}{2} - \theta \right)\)
-
\(\dfrac{\sec(-\alpha) - 1}{\sec(\alpha) + 1} = \dfrac{\sin^2(-\alpha)}{(\cos(-\alpha) + 1)^2}\)
-
\(\sec(-6t) = \sec(6t)\)
-
\(-\dfrac{\cos^2(-A)}{1 + \sin(-A)} = -1 + \sin(-A)\)
-
\( \dfrac{\tan\left( \theta \right) + \tan\left( \phi \right)}{\cot\left( \theta \right) + \cot\left( \phi \right)} = \tan\left( \theta \right) \tan\left( \phi \right) \)
-
\(\dfrac{\sin^2(-\alpha)}{\cos(-\alpha)} = -\tan(\alpha) \cdot \sin(-\alpha)\)
-
\(\dfrac{\sec(y) - 1}{\sec(y) + 1} = \dfrac{1 - \cos(-y)}{1 + \cos(-y)}\)
-
\(\dfrac{\sin^2\left( \frac{\pi}{2} - A \right)}{1 - \sin(A)} = \dfrac{\cos^2\left( A \right)}{1 - \cos\left(\frac{\pi}{2} - A\right)}\)
-
\(\dfrac{1 + \sin\left( \frac{\pi}{2} - B \right)}{\sin(B)} = \sec\left( \frac{\pi}{2} - B \right) + \cot(B)\)
-
\(\dfrac{\sin\left( \frac{\pi}{2} - \phi \right) - 1}{\sin\left( \frac{\pi}{2} - \phi \right)} = \dfrac{\sin^2(\phi)}{\sin\left( \frac{\pi}{2} - \phi \right)\left(1 + \sin\left( \frac{\pi}{2} - \phi \right)\right)}\)
-
\(\cot(9 - 7\theta) = -\cot(7\theta - 9)\)
-
\(\dfrac{\csc\left( \frac{\pi}{2} - B \right) - 1}{\sec(B) + 1} = \dfrac{\cos^2\left( \frac{\pi}{2} - B \right)}{\left(\sin\left( \frac{\pi}{2} - B \right) + 1\right)^2}\)
-
\(\dfrac{1 - \cos(\alpha)}{\sin^2(\alpha)} = \dfrac{1}{1 + \cos(\alpha)}\)
Challenge Problems
-
Suppose that you are given a system of two equations of the following form:\[\begin{array}{rcl}
A \cos\left(\phi\right) & = & B \nu_1 - B\nu_2 \cos\left(\theta\right) \\[6pt] A\sin\left(\phi\right) & = & B \nu_2 \sin\left(\theta\right) \\[6pt] \end{array} \nonumber \]Show that\[A^2 = B^2 \left( \nu_1^2 + \nu_2^2 - 2\nu_1 \nu_2 \cos\left(\theta\right) \right).\nonumber \] -
Suppose that a point with coordinates\[\left(x,y\right) = \left(a \left(\cos\left(\psi\right) - \epsilon\right), a\sqrt{1 - \epsilon^2} \sin\left(\psi\right)\right) \nonumber \]is a distance \(r>0 \) from the origin, where \(a>0 \) and \(0 < \epsilon < 1 \). Use \(r^2 = x^2 + y^2\) to show that \(r = a\left(1 - \epsilon \cos\left(\psi\right) \right)\).
(Note: These coordinates arise in the study of elliptical orbits of planets.) -
Show that each trigonometric function can be written solely in terms of the sine function.
This page titled 9.1.2: Homework was last modified on Tue, 08 Jul 2025 17:45:37 GMT and is shared under a not declared license and was authored, remixed, and/or curated by Roy Simpson.
The book could not be loaded just now.