# trigomatic

• Jun 7th 2006, 05:53 AM
watsonmath
trigomatic
Help on this please!!! Need working or explaination for these questions.
*** Answer given but some might be wrong. THANKS!

change to an expression containing only sin and cos.

1
a.tan0csc0 Ans:1/cos0
b. sec0-tan0sin0 Ans:cos0
c.(tan0/csc0)+(sin/tan0) Ans:1/cos0
d. cot0+tan0 Ans:1/(sin0cos0)
e. (sec^sqrt0-1)/(sin^sqrt0) Ans:1/cos^sqrt0
f. (cot^sqrt0)/(csc^sqrt0) Ans:cos^sqrt0

2. Simplify
a. (sin x/csc x)+ (cos x/sec x) Ans:1
b.1/cot0sec0 Ans:sin0
• Jun 7th 2006, 07:35 AM
Soroban
Hello, watsonmath!

You really can't do any of these??

Okay, I must assume you know the fundamental identities:

. . $\displaystyle \csc\theta = \frac{1}{\sin\theta}\qquad\sec\theta = \frac{1}{\cos\theta}$ . . . $\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta}\qquad\cot\theta = \frac{\cos\theta}{\sin\theta}$

$\displaystyle \sin^2\theta + \cos^2\theta \:=\:1\quad\Rightarrow\quad \sin^2\theta \:=\:1 - \cos^2\theta$ . . . $\displaystyle \Rightarrow\quad\cos^2\theta\:=\:1 - \sin^2\theta$

Quote:

1. Change to an expression containing only sin and cos.

$\displaystyle a)\;\tan\theta\csc\theta$
We have: $\displaystyle \tan\theta\csc\theta\:=\:\frac{\sin\theta}{\cos \theta}\cdot\frac{1}{\sin\theta}\:=\:\frac{1}{\cos \theta}$

Quote:

$\displaystyle b)\;\sec\theta - \tan\theta\sin\theta$
$\displaystyle \sec\theta - \tan\theta\sin\theta \;=\;\frac{1}{\cos\theta} -\frac{\sin\theta}{\cos\theta}\sin\theta \;=$ .$\displaystyle \frac{1 - \sin^2\theta}{\cos\theta}\;=\;\frac{\cos^2\theta}{ \cos\theta} \;=\;\cos\theta$

Quote:

$\displaystyle c)\; \frac{\tan\theta}{\csc\theta} + \frac{\sin\theta}{\tan\theta}$
We have: $\displaystyle \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\sin \theta}} + \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}}\; =\;\frac{\sin^2\theta}{\cos\theta} + \cos\theta \;=\;\frac{\sin^2\theta + \cos^2\theta}{\cos\theta} \;= \;\frac{1}{\cos\theta}$

Quote:

$\displaystyle d)\;\cot\theta + \tan\theta$
We have: $\displaystyle \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} \;=\;\frac{\cos^2\theta}{\sin\theta\cos\theta} + \frac{\sin^2\theta}{\sin\theta\cos\theta}\;=$ $\displaystyle \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \;=\;\frac{1}{\sin\theta\cos\theta}$

Quote:

$\displaystyle e)\;\frac{\sec^2\theta -1}{\sin^2\theta}$
We have: $\displaystyle \frac{\tan^2\theta}{\sin^2\theta}\;=\;\frac{\frac{ \sin^2\theta}{\cos^2\theta}}{\sin^2\theta} \;=\;\frac{1}{\cos^2\theta}$

Quote:

$\displaystyle f)\;\frac{\cot^2\theta}{\csc^2\theta}$
We have: $\displaystyle \frac{\frac{\cos^2\theta}{\sin^2\theta}}{\frac{1}{ \sin^2\theta}}\;=\;\cos^2\theta$

Quote:

2. Simplify

$\displaystyle a)\;\frac{\sin x}{\csc x} + \frac{\cos x}{\sec x}$
We have: $\displaystyle \frac{\sin x}{\frac{1}{\sin x}} + \frac{\cos x}{\frac{1}{\cos x}} \;=\;\sin^2x + \cos^2x\;=\;1$

Quote:

$\displaystyle b)\;\frac{1}{\cot\theta\sec\theta}$
We have: $\displaystyle \frac{1}{\frac{\cos\theta}{\sin\theta}\cdot\frac{1 }{\cos\theta}} \;=\;\frac{1}{\frac{1}{\sin\theta}} \;=\;\sin\theta$