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Find the value of each of the following: $\newcommand {\arccos}{\textrm{arccos}\,}$ $\newcommand {\arcsin}{\textrm{arcsin}\,}$ $\newcommand {\arctan}{\textrm{arctan}\,}$ $\newcommand {\arcsec}{\textrm{arcsec}\,}$ $\newcommand {\arccsc}{\textrm{arccsc}\,}$ $\newcommand {\arccot}{\textrm{arccot}\,}$
$\tan \left( -\frac{7\pi}{6} \right)$
$\cot \theta, \textrm{ if } \sin \theta = \frac{2}{3} \textrm{ and } \frac{\pi}{2} \lt \theta \lt \pi$
$\cos \left( \frac{-11\pi}{6} \right)$
$\csc t, \textrm{ if } \cos t = -\frac{15}{17} \textrm{ and } \pi \lt t \lt \frac{3\pi}{2}$
$\tan \left(-\frac{\pi}{6} \right)$
$\arccos \left( \sin \frac{11\pi}{4} \right)$
$\cos(\arctan(-1))$
$\sec(\arccot(\frac{-5}{12}))$
$\arcsec(-\sqrt{2})$
$\sec(\arccot \frac{x}{3})$
$\arctan(\sin(-\frac{5\pi}{2}))$
$\sec(-\frac{5\pi}{6})$
$9\arccot^2 \frac{\sqrt{3}}{3}$
$\arccsc(-2)$
$\tan(\arcsin(-\frac{3}{5}))$
$\arccos(\cot \frac{11\pi}{4})$
$\csc t, \textrm{ if } \tan t = 3 \textrm{ and } \pi \lt t \lt \frac{3\pi}{2}$
$\arccsc(\cos(\frac{-\pi}{3}))$
$\tan \theta, \textrm{ if } \sin \theta = \frac{5}{13} \textrm{ and } \frac{\pi}{2} \lt \theta \lt \pi$
$\tan \frac{9\pi}{2}$
$\cos(\arctan \frac{x}{3})$
$\arcsin(\cos \frac{\pi}{2})$
$\arcsec 1$
$\csc(-\frac{9\pi}{4})$
$\cos(\arcsin x)$
$\tan \frac{54\pi}{6}$
See full solutions.
$-\frac{\sqrt{3}}{3}$
$-\frac{\sqrt{5}}{2}$
$\frac{\sqrt{3}}{2}$
$-\frac{17}{8}$
$-\frac{\sqrt{3}}{3}$
$\frac{\pi}{4}$
$\frac{\sqrt{2}}{2}$
$-\frac{13}{5}$
$\frac{3\pi}{4}$
$\frac{\sqrt{x^2+9}}{x}$
$-\frac{\pi}{4}$
$-\frac{2}{\sqrt{3}}$
$\pi^2$
$-\frac{\pi}{6}$
$-\frac{3}{4}$
$\pi$
$-\frac{\sqrt{10}}{3}$
no value
$-\frac{5}{12}$
no value
$\frac{3}{\sqrt{x^2 + 9}}$
$0$
$0$
$-\sqrt{2}$
$\sqrt{1-x^2}$
$0$
Find the value of each of the following: $\newcommand {\arccos}{\textrm{arccos}\,}$ $\newcommand {\arcsin}{\textrm{arcsin}\,}$ $\newcommand {\arctan}{\textrm{arctan}\,}$ $\newcommand {\arcsec}{\textrm{arcsec}\,}$ $\newcommand {\arccsc}{\textrm{arccsc}\,}$ $\newcommand {\arccot}{\textrm{arccot}\,}$
$\arcsec^2 (\csc \frac{2\pi}{3})$
$\cos^2 \frac{7\pi}{6}$
$\arccsc(-\sqrt{2})$
$5\arccot(-\frac{\sqrt{3}}{3})$
$\cot(\arcsin(-\frac{8}{17}))$
$\arcsin(\sin \frac{5\pi}{3})$
$\cos(\arctan 5)$
$\arccos(\sin(-\frac{\pi}{6}))$
$\sin t, \textrm{ if } \cot t = -\frac{12}{5} \textrm{ and } \frac{3\pi}{2} \lt t \lt 2\pi$
$\cos(\arctan(\frac{-3}{4}))$
$\arcsec(-\frac{2\sqrt{3}}{3})$
$\arccos(\sin \frac{23\pi}{4})$
$\cot t, \textrm{ if } \sec t = -\frac{8}{5} \textrm{ and } \pi \lt t \lt \frac{3\pi}{2}$
See full solutions.
$\frac{\pi^2}{36}$
$\frac{3}{4}$
$-\frac{\pi}{4}$
$\frac{2\pi}{3}$
$-\frac{15}{8}$
$-\frac{\pi}{3}$
$\frac{\sqrt{26}}{26}$
$\frac{2\pi}{3}$
$-\frac{5}{13}$
$\frac{4}{5}$
$\frac{5\pi}{6}$
$\frac{3\pi}{4}$
$\frac{5\sqrt{39}}{39}$
Sketch graphs of the following. Label intercepts, asymptotes, and endpoints.
$y = -4\cos(\frac{x}{2} - \frac{\pi}{4})$ on $[-\pi,3\pi]$
$y = 4\sin(\frac{2}{3} x + \frac{\pi}{6})$ from $-2\pi$ to $2\pi$
$y = -2\sec(3x+\pi)$ on $[-\frac{\pi}{3},\frac{2\pi}{3}]$
$y = 2\csc(2x + \frac{\pi}{2})$ for $-\frac{\pi}{2} \le x \le \pi$
$y = 5\cos(3x + \frac{\pi}{2})$ on $[0,\pi]$
$y = 4 - 2\sin(\frac{x}{3}-\frac{\pi}{3})$ from $-2\pi$ to $\frac{5\pi}{2}$
$y = \tan(x + \frac{\pi}{4})$ on $[-\pi,2\pi]$
$y = -2\sin(x - \frac{3\pi}{2})$ on $[-\frac{\pi}{2},2\pi]$
$y = \cot(x - \frac{\pi}{6})$ from $-\pi$ to $2\pi$
$y = -\frac{1}{3} \cos(\frac{x}{2} + \frac{\pi}{4})$ on $[-2\pi,4\pi]$
$y = -\sec \frac{x}{2}$ for $-3\pi \lt x \lt 3\pi$
See full solutions.
Show whether each of the following is or is not an identity:
$\displaystyle{\frac{1}{\csc \theta + \cot \theta} + \frac{\sec \theta + 1}{\tan \theta} = 2\csc \theta}$
$\displaystyle{\frac{1+\tan \theta}{1-\tan \theta} + \frac{1 + \cot \theta}{1 - \cot \theta} = 0}$
$\displaystyle{\frac{1}{1+\cos \theta} - \frac{1}{1-\cos \theta} = \frac{2}{\sec \theta - \cos \theta}}$
$\displaystyle{(1+\sec x)(1-\cos x) = \tan x \sin x}$
$\displaystyle{\sec 2\theta = \frac{\sec^2 \theta}{2 - \sec^2 \theta}}$
$\displaystyle{\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} = 2 \sec \theta}$
$\displaystyle{\frac{\tan^2 x \csc^2 x - 1}{\csc x \tan^2 x \sin x} = 1}$
$\displaystyle{\frac{1}{\sec \theta - \tan \theta} = \sec \theta + \tan \theta}$
$\displaystyle{\frac{\sin \theta + \tan \theta}{1 + \cos \theta} = \cot \theta}$
$\displaystyle{\cos^4 x - \sin^4 x = \cos 2x}$
$\displaystyle{\frac{\tan t}{\tan^2 t - 1} = \frac{1}{\tan t - \cot t}}$
$\displaystyle{\sec x - \sin x \tan x = \cos x}$
$\displaystyle{\frac{\sin \theta + \cos \theta}{\sec \theta + \csc \theta} = \frac{\sin \theta}{\sec \theta}}$
$\displaystyle{\frac{\cos^4 x - \sin^4 x}{\sin x + \cos x} = \frac{1-\tan x}{1+\tan x}}$
Note, (c), (i), (n) are not identities; the others are identities
See full solutions.
Find all solutions of the following equations:
$\displaystyle{\tan^2 x + \sec^2 x + 3\sec x = 1}$
$\displaystyle{\cos 2x = 1 + \sin x}$
$\displaystyle{2\sin x \tan x + \tan x - 2\sin x - 1 = 0}$
$\displaystyle{2\cos^2 x - 3\cos x + 1 = 0}$
$\displaystyle{3\tan x + \frac{1}{\tan x} = 2\sqrt{3}}$
$\displaystyle{2\sin x \tan x = 3}$
$\displaystyle{\sec^2 x - \tan x = 1}$
$\displaystyle{4 \sin^2 x - 1 = 0}$
$\displaystyle{\cos 2x = \cos x}$
$\displaystyle{2\cos^3 x + \sin^2 x = 1}$
$\displaystyle{8\sin^4 x - 10 \sin^2 x + 3 = 0}$
$\displaystyle{2\sin^2 x + 7\sin x + 3 = 0}$
$\displaystyle{3 \cot x = \tan x}$
See the full solutions.
Note: the condition that $n=0,1,2,3,\ldots$ should be considered attached to all answers below:
$\frac{2\pi}{3} \pm 2\pi n; \frac{4\pi}{3} \pm 2\pi n$
$0 \pm \pi n; \frac{7\pi}{6} \pm \pi n; \frac{11\pi}{6} \pm \pi n$
$\frac{7\pi}{6} \pm 2\pi n; \frac{11\pi}{6} \pm 2\pi n; \frac{\pi}{4} \pm \pi n$
$\frac{\pi}{3} \pm 2\pi n; \frac{5\pi}{3} \pm 2\pi n; 0 \pm 2\pi n$
$\frac{\pi}{6} \pm \pi n$
$\frac{\pi}{3} \pm 2\pi n; \frac{5\pi}{3} \pm 2\pi n$
$0 \pm \pi n; \frac{\pi}{4} \pm \pi n$
$\frac{\pi}{6} \pm \pi n; \frac{5\pi}{6} \pm \pi n$
$\frac{2\pi}{3} \pm 2\pi n; \frac{4\pi}{3} \pm 2\pi n; 0 \pm 2\pi n$
$\frac{\pi}{2} \pm \pi n; \frac{\pi}{3} \pm 2\pi n; \frac{5\pi}{3} \pm 2\pi n$
$\frac{\pi}{4} \pm \pi n; \frac{3\pi}{4} \pm \pi n; \frac{\pi}{3} \pm \pi n; \frac{2\pi}{3} \pm \pi n$
$\frac{7\pi}{6} \pm 2\pi n; \frac{11\pi}{6} \pm 2\pi n$
$\frac{\pi}{3} \pm \pi n; \frac{2\pi}{3} \pm \pi n$
Verify the following identities:
$\displaystyle{\frac{1}{\tan x + \cot x} = (\sin x)(\cos x)}$
$\displaystyle{\sec^2 \alpha + \csc^2 \alpha = \tan^2 \alpha + \cot^2 \alpha + 2}$
$\displaystyle{\frac{\tan x - \sin x}{\tan x + \sin x} = \frac{\sec x -1}{\sec x + 1}}$
$\displaystyle{(1 - \cos^2 \theta)(1+ \cot^2 \theta) = 1}$
$\displaystyle{\frac{\csc \beta - \sin \beta}{1 - \sin^2 \beta} = \csc \beta}$
$\displaystyle{\frac{\sin \gamma \sec^2 \gamma - \sin \gamma}{\cos \gamma} = \tan^3 \gamma}$
$\displaystyle{\frac{\cot^2 \theta - 1}{1 - \tan^2 \theta} = \cot^2 \theta}$
$\displaystyle{\frac{\tan \alpha}{\sec \alpha + 1} = \frac{1}{\cot \alpha + \csc \alpha}}$
$\displaystyle{\frac{\cos^2 x + 3\cos x + 2}{\sin^2 x} = \frac{2 + \cos x}{1 - \cos x}}$
$\displaystyle{\frac{1+\sin x + \cos x}{1+\cos x - \sin x} = \sec x + \tan x}$ (challenge!)
See full solutions.