Chapter 9

All three functions, $F$,$G$,$$ and $H$,$$ are even.

This is because $F\left(-x\right)=\sin \left(-x\right)\sin \left(-x\right)=\left(-\sin x\right)\left(-\sin x\right)={\sin }^{2}x=F\left(x\right)$,$G\left(-x\right)=\cos \left(-x\right)\cos \left(-x\right)=\cos x\cos x={\cos }^{2}x=G\left(x\right)$ and $H\left(-x\right)=\tan \left(-x\right)\tan \left(-x\right)=\left(-\tan x\right)\left(-\tan x\right)={\tan }^{2}x=H\left(x\right).$

When $\cos t=0,$ then $\sec t=\frac{1}{0},$ which is undefined.

$\sin x$

$\sec x$

$\csc t$

$\mathrm{-1}$

${\sec }^{2}x$

${\sin }^{2}x+1$

$\frac{1}{\sin x}$

$\frac{1}{\cot x}$

$\tan x$

$-4\sec x\tan x$

$\pm \sqrt{\frac{1}{{\cot }^{2}x}+1}$

$\frac{\pm \sqrt{1-{\sin }^{2}x}}{\sin x}$

Answers will vary. Sample proof:

$\begin{array}{ccc}\hfill \cos x-{\cos }^{3}x& =& \cos x(1-{\cos }^{2}x)\hfill \\ & =& \cos x{\sin }^{2}x\hfill \end{array}$

Answers will vary. Sample proof:
$\frac{1+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}={\sec }^{2}x+{\tan }^{2}x={\tan }^{2}x+1+{\tan }^{2}x=1+2{\tan }^{2}x$

Answers will vary. Sample proof:
${\cos }^{2}x-{\tan }^{2}x=1-{\sin }^{2}x-\left({\sec }^{2}x-1\right)=1-{\sin }^{2}x-{\sec }^{2}x+1=2-{\sin }^{2}x-{\sec }^{2}x$

False

False

Proved with negative and Pythagorean identities

True $3{\sin }^2\theta +4{\cos }^2\theta =3{\sin }^2\theta +3{\cos }^2\theta +{\cos }^2\theta =3\left({\sin }^2\theta +{\cos }^2\theta \right)+{\cos }^2\theta =3+{\cos }^2\theta$

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures $x,$ the second angle measures $\frac{\pi }{2}-x.$ Then $\sin x=\cos \left(\frac{\pi }{2}-x\right).$ The same holds for the other cofunction identities. The key is that the angles are complementary.

$\sin \left(-x\right)=-\sin x,$ so $\sin x$ is odd. $\cos \left(-x\right)=\cos \left(0-x\right)=\cos x,$ so $\cos x$ is even.

$\frac{\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

$-2-\sqrt{3}$

$-\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x$

$-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x$

$\csc \theta$

$\cot x$

$\tan \left(\frac{x}{10}\right)$

$\begin{array}{ccccc}\hfill \sin (a-b)& =& \left(\frac{4}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{3}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)\hfill & =& \frac{4-6\sqrt{2}}{15}\hfill \\ \hfill \cos (a+b)& =& \left(\frac{3}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{4}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)\hfill & =& \frac{3-8\sqrt{2}}{15}\hfill \end{array}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$

$\sin x$

$\cot \left(\frac{\pi }{6}-x\right)$

$\cot \left(\frac{\pi }{4}+x\right)$

$\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}$

They are the same.

They are the different, try $g\left(x\right)=\sin \left(9x\right)-\cos \left(3x\right)\sin \left(6x\right).$

They are the same.

They are the different, try $g\left(\theta \right)=\frac{2\tan \theta }{1-{\tan }^2\theta }.$

They are different, try $g\left(x\right)=\frac{\tan x-\tan \left(2x\right)}{1+\tan x\tan \left(2x\right)}.$

$-\frac{\sqrt{3}-1}{2\sqrt{2}},\text{or }-0.2588$

$\frac{1+\sqrt{3}}{2\sqrt{2}},$ or 0.9659

$\begin{array}{ccc}\hfill \tan \left(x+\frac{\pi }{4}\right)& =& \\ \hfill \frac{\tan x+\tan \left(\frac{\pi }{4}\right)}{1-\tan x\tan \left(\frac{\pi }{4}\right)}& =& \\ \hfill \frac{\tan x+1}{1-\tan x(1)}& =& \frac{\tan x+1}{1-\tan x}\hfill \end{array}$

$\begin{array}{ccc}\hfill \frac{\cos (a+b)}{\cos a\cos b}& =& \\ \hfill \frac{\cos a\cos b}{\cos a\cos b}-\frac{\sin a\sin b}{\cos a\cos b}& =& 1-\tan a\tan b\hfill \end{array}$

$\begin{array}{ccc}\hfill \frac{\cos (x+h)-\cos x}{h}& =& \\ \hfill \frac{\cos x\cosh -\sin x\sinh -\cos x}{h}& =& \\ \hfill \frac{\cos x(\cosh -1)-\sin x\sinh }{h}& =& \cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\hfill \end{array}$

True

True. Note that $\sin \left(\alpha +\beta \right)=\sin \left(\pi -\gamma \right)$ and expand the right hand side.

Use the Pythagorean identities and isolate the squared term.

$\frac{1-\cos x}{\sin x},\frac{\sin x}{1+\cos x},$ multiplying the top and bottom by $\sqrt{1-\cos x}$ and $\sqrt{1+\cos x},$ respectively.

a) $\frac{3\sqrt{7}}{32}$ b) $\frac{31}{32}$ c) $\frac{3\sqrt{7}}{31}$

a) $\frac{\sqrt{3}}{2}$ b) $-\frac{1}{2}$ c) $-\sqrt{3}$

$\cos \theta =-\frac{2\sqrt{5}}{5},\sin \theta =\frac{\sqrt{5}}{5},\tan \theta =-\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =-\frac{\sqrt{5}}{2},\cot \theta =-2$

$2\sin \left(\frac{\pi }{2}\right)$

$\frac{\sqrt{2-\sqrt{2}}}{2}$

$\frac{\sqrt{2-\sqrt{3}}}{2}$

$2+\sqrt{3}$

$-1-\sqrt{2}$

a) $\frac{3\sqrt{13}}{13}$ b) $-\frac{2\sqrt{13}}{13}$ c) $-\frac{3}{2}$

a) $\frac{\sqrt{10}}{4}$ b) $\frac{\sqrt{6}}{4}$ c) $\frac{\sqrt{15}}{3}$

$\frac{120}{169},–\frac{119}{169},–\frac{120}{119}$

$\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13},\frac{2}{3}$

$\cos (\mathrm{74°})$

$\cos (18x)$

$3\sin (10x)$

$-2\sin \left(-x\right)\cos \left(-x\right)=-2(-\sin \left(x\right)\cos \left(x\right))=\sin \left(2x\right)$

$\begin{array}{ccc}\hfill \frac{\sin (2\theta )}{1+\cos (2\theta )}{\tan }^2\theta & =& \frac{2\sin (\theta )\cos (\theta )}{1+{\cos }^2\theta -{\sin }^2\theta }{\tan }^2\theta =\hfill \\ \hfill \frac{2\sin (\theta )\cos (\theta )}{2{\cos }^2\theta }{\tan }^2\theta & =& \frac{\sin (\theta )}{\cos \theta }{\tan }^2\theta =\hfill \\ \hfill \cot (\theta ){\tan }^2\theta & =& {\tan }^3\theta \hfill \end{array}$

$\frac{1+\cos (12x)}{2}$

$\frac{3+\cos (12x)-4\cos (6x)}{8}$

$\frac{2+\cos (2x)-2\cos (4x)-\cos (6x)}{32}$

$\frac{3+\cos (4x)-4\cos (2x)}{3+\cos (4x)+4\cos (2x)}$

$\frac{1-\cos (4x)}{8}$

$\frac{3+\cos (4x)-4\cos (2x)}{4(\cos (2x)+1)}$

$\frac{\left(1+\cos \left(4x\right)\right)\sin x}{2}$

$4\sin x\cos x\left({\cos }^{2}x-{\sin }^{2}x\right)$

$\begin{array}{l}\frac{2\tan x}{1+{\tan }^{2}x}=\frac{\frac{2\sin x}{\cos x}}{1+\frac{{\sin }^{2}x}{{\cos }^{2}x}}=\frac{\frac{2\sin x}{\cos x}}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}=\\ \frac{2\sin x}{\cos x}.\frac{{\cos }^{2}x}{1}=2\sin x\cos x=\sin (2x)\end{array}$

$\frac{2\sin x\cos x}{2{\cos }^{2}x-1}=\frac{\sin (2x)}{\cos (2x)}=\tan (2x)$

$\begin{array}{ccc}\hfill \sin (x+2x)& =& \sin x\cos (2x)+\sin (2x)\cos x\hfill \\ & =& \sin x\left({\cos }^{2}x-{\sin }^{2}x\right)+2\sin x\cos x\cos x\hfill \\ & =& \sin x{\cos }^{2}x-{\sin }^{3}x+2\sin x{\cos }^{2}x\hfill \\ & =& 3\sin x{\cos }^{2}x-{\sin }^{3}x\hfill \end{array}$

$\begin{array}{ccc}\hfill \frac{1+\cos (2t)}{\sin (2t)-\cos t}& =& \frac{1+2{\cos }^{2}t-1}{2\sin t\cos t-\cos t}\hfill \\ & =& \frac{2{\cos }^{2}t}{\cos t(2\sin t-1)}\hfill \\ & =& \frac{2\cos t}{2\sin t-1}\hfill \end{array}$

$\begin{array}{ccc}\hfill \left({\cos }^2(4x)-{\sin }^2(4x)-\sin (8x))({\cos }^2(4x)-{\sin }^2(4x)+\sin (8x)\right)& =& \\ & =& (\cos (8x)-\sin (8x))(\cos (8x)+\sin (8x))\hfill \\ & =& {\cos }^2(8x)-{\sin }^2(8x)\hfill \\ & =& \cos (16x)\hfill \end{array}$