Simplifying trigonometric functions

**aikenfan** · November 18th 2007, 06:07 PM

Where do I go on a problem like this? I am supposed to simplify them...I have done about 10 of these earlier today, but I am stuck on these last two...I know that there is probably more than one way to go on these, but I am having a hard time getting started

**aikenfan** · November 18th 2007, 06:10 PM

for 66, I have done:
Common denominator

(Sec x - 1) - (sec x + 1)/ (Sec x - 1)(sec x + 1)

**Jhevon** · November 18th 2007, 06:15 PM

Originally Posted by aikenfan

for 66, I have done:
Common denominator

(Sec x - 1) + (sec x + 1)/ (Sec x - 1)(sec x + 1)

that is wrong. you were on the right track, but you messed up the signs

$\frac 1{\sec x + 1} - \frac 1{\sec x - 1} = \frac {\sec x - 1 - \sec x - 1}{(\sec x + 1)(\sec x - 1)}$

note that the denominator is the difference of two squares, thus we have:

$\frac {-2}{\sec^2 - 1}$

now what?

continue the same way for the next problem. you don't have many options besides combining the fractions, follow your nose

**aikenfan** · November 18th 2007, 06:21 PM

this would make it -2/tan^2x
because of Pythagorean Identities: tan^2x = sec^2x - 1

**Jhevon** · November 18th 2007, 06:24 PM

Originally Posted by aikenfan

this would make it -2/tan^2x
because of Pythagorean Identities: tan^2x = sec^2x - 1

yes. but what is 1/tan^2(x) ?

**aikenfan** · November 18th 2007, 06:25 PM

cot^2x

**Jhevon** · November 18th 2007, 06:26 PM

Originally Posted by aikenfan

cot^2x

yes, so we have -2cot^2(x)

now, move on to the second

**aikenfan** · November 18th 2007, 06:28 PM

So far, I have got:
(cosx(1 + sinx) + 1 + sinx(1 + sinx))/(1 + sinx)(cosx)
I am guessing that I have to distribute these

**Jhevon** · November 18th 2007, 06:31 PM

Originally Posted by aikenfan

So far, I have got:
(cosx(1 + sinx) + 1 + sinx(1 + sinx))/(1 + sinx)(cosx)
I am guessing that I have to distribute these

yup, that would be a good idea. again, you don't have many other options, so just follow your nose

**aikenfan** · November 18th 2007, 06:34 PM

I get:
(cosx + cosx(sinx)) + (1 + sinx + sinx + sinx^2)/ (1 + sinx)(cosx)

= (cosx + cosx(sinx)) + (1 + 2sinx + sinx^2)/ (1 + sinx)(cosx)

**Jhevon** · November 18th 2007, 06:40 PM

Originally Posted by aikenfan

I get:
(cosx + cosx(sinx)) + (1 + sinx + sinx + sinx^2)/ (1 + sinx)(cosx)

= (cosx + cosx(sinx)) + (1 + 2sinx + sinx^2)/ (1 + sinx)(cosx)

where did cosx(sinx) come from, you shouldn't have that

**Soroban** · November 18th 2007, 06:45 PM

Hello, aikenfan!

There is a trick we can use on the second problem . . .

$68)\;\;\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}$

Multiply the first fraction by $\frac{1-\sin x}{1-\sin x}$

. . We have: . $\frac{\cos x}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} \;\;=\;\;\frac{\cos x(1 -\sin x)}{1-\sin^2\!x} \;\;=\;\;\frac{\cos x(1-\sin x)}{\cos^2\!x} \;\;=\;\;\frac{1-\sin x}{\cos x}$

The problem becomes: . $\frac{1-\sin x}{\cos x} + \frac{1+\sin x}{\cos x} \;\;=\;\;\frac{2}{\cos x} \;\;=\;\;2\sec x$

**aikenfan** · November 18th 2007, 06:53 PM

Thank you both very much for all of your time and help!

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