trigomatic

**watsonmath** · June 7th 2006, 05:53 AM

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

**Soroban** · June 7th 2006, 07:35 AM

Hello, watsonmath!

You really can't do any of these??

Okay, I must assume you know the fundamental identities:

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

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

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

$a)\;\tan\theta\csc\theta$

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

$b)\;\sec\theta - \tan\theta\sin\theta$

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

$c)\; \frac{\tan\theta}{\csc\theta} + \frac{\sin\theta}{\tan\theta}$

We have: $\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\sin \theta}} + \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}}\; =\;\frac{\sin^2\theta}{\cos\theta} + \cos\theta \;=\;\frac{\sin^2\theta + \cos^2\theta}{\cos\theta} \;= \;\frac{1}{\cos\theta}$

$d)\;\cot\theta + \tan\theta$

We have: $\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} \;=\;\frac{\cos^2\theta}{\sin\theta\cos\theta} + \frac{\sin^2\theta}{\sin\theta\cos\theta}\;=$ $\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \;=\;\frac{1}{\sin\theta\cos\theta}$

$e)\;\frac{\sec^2\theta -1}{\sin^2\theta}$

We have: $\frac{\tan^2\theta}{\sin^2\theta}\;=\;\frac{\frac{ \sin^2\theta}{\cos^2\theta}}{\sin^2\theta} \;=\;\frac{1}{\cos^2\theta}$

$f)\;\frac{\cot^2\theta}{\csc^2\theta}$

We have: $\frac{\frac{\cos^2\theta}{\sin^2\theta}}{\frac{1}{ \sin^2\theta}}\;=\;\cos^2\theta$

2. Simplify

$a)\;\frac{\sin x}{\csc x} + \frac{\cos x}{\sec x}$

We have: $\frac{\sin x}{\frac{1}{\sin x}} + \frac{\cos x}{\frac{1}{\cos x}} \;=\;\sin^2x + \cos^2x\;=\;1$

$b)\;\frac{1}{\cot\theta\sec\theta}$

We have: $\frac{1}{\frac{\cos\theta}{\sin\theta}\cdot\frac{1 }{\cos\theta}} \;=\;\frac{1}{\frac{1}{\sin\theta}} \;=\;\sin\theta<br />$

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