9.1 Solving trigonometric equations with identities  (Page 5/9)

We know $g(x)=\cos x$ is an even function, and $f(x)=\sin x$ and $h(x)=\tan x$ are odd functions. What about $G(x)={\cos }^{2}x,F(x)={\sin }^{2}x,$ and $H(x)={\tan }^{2}x?$ Are they even, odd, or neither? Why?

Examine the graph of $f(x)=\sec x$ on the interval $[-\pi ,\pi ].$ How can we tell whether the function is even or odd by only observing the graph of $f(x)=\sec x?$

After examining the reciprocal identity for $\sec t,$ explain why the function is undefined at certain points.

All of the Pythagorean identities are related. Describe how to manipulate the equations to get from ${\sin }^{2}t+{\cos }^{2}t=1$ to the other forms.

$\sin x\cos x\sec x$

$\sin (-x)\cos (-x)\csc (-x)$

$\tan x\sin x+\sec x{\cos }^{2}x$

$\csc x+\cos x\cot (-x)$

$\frac{\cot t+\tan t}{\sec (-t)}$

$3{\sin }^{3}t\csc t+{\cos }^{2}t+2\cos (-t)\cos t$

$-\tan (-x)\cot (-x)$

$\frac{-\sin (-x)\cos x\sec x\csc x\tan x}{\cot x}$

$\frac{1+{\tan }^2\theta }{{\csc }^2\theta }+{\sin }^2\theta +\frac{1}{{\sec }^2\theta }$

$\left(\frac{\tan x}{{\csc }^{2}x}+\frac{\tan x}{{\sec }^{2}x}\right)\left(\frac{1+\tan x}{1+\cot x}\right)-\frac{1}{{\cos }^{2}x}$

$\frac{1-{\cos }^{2}x}{{\tan }^{2}x}+2{\sin }^{2}x$

$\frac{\tan x+\cot x}{\csc x};\cos x$

$\frac{\sec x+\csc x}{1+\tan x};\sin x$

$\frac{\cos x}{1+\sin x}+\tan x;\cos x$

$\frac{1}{\sin x\cos x}-\cot x;\cot x$

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x};\csc x$

$\left(\sec x+\csc x\right)\left(\sin x+\cos x\right)-2-\cot x;\tan x$

$\frac{1}{\csc x-\sin x};\sec x\text{ and }\tan x$

$\frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x};\sec x\text{ and }\tan x$

$\tan x;\sec x$

$\sec x;\cot x$

$\sec x;\sin x$

$\cot x;\sin x$

$\cot x;\csc x$

$\cos x-{\cos }^{3}x=\cos x{\sin }^{2}x$

$\cos x\left(\tan x-\sec \left(-x\right)\right)=\sin x-1$

$\frac{1+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}=1+2{\tan }^{2}x$

${\left(\sin x+\cos x\right)}^2=1+2\sin x\cos x$

${\cos }^{2}x-{\tan }^{2}x=2-{\sin }^{2}x-{\sec }^{2}x$

$\frac{1}{1+\cos x}-\frac{1}{1-\cos (-x)}=-2\cot x\csc x$

${\csc }^{2}x\left(1+{\sin }^{2}x\right)={\cot }^{2}x$

$\left(\frac{{\sec }^2(-x)-{\tan }^{2}x}{\tan x}\right)\left(\frac{2+2\tan x}{2+2\cot x}\right)-2{\sin }^{2}x=\cos 2x$

$\frac{\tan x}{\sec x}\sin \left(-x\right)={\cos }^{2}x$

$\frac{\sec \left(-x\right)}{\tan x+\cot x}=-\sin \left(-x\right)$

$\frac{1+\sin x}{\cos x}=\frac{\cos x}{1+\sin \left(-x\right)}$

$\frac{{\cos }^2\theta -{\sin }^2\theta }{1-{\tan }^2\theta }={\sin }^2\theta$

$3{\sin }^2\theta +4{\cos }^2\theta =3+{\cos }^2\theta$

$\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }={\sec }^2\theta$