# Substitute x= 3 sec(theta) and simplify

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• Dec 7th 2012, 04:16 PM
mathisfun26
Substitute x= 3 sec(theta) and simplify
$\displaystyle \frac{\sqrt{x^{2}+9}}{x^{2}}$. Help write in terms of sines and cosines can anyone give me steps on how to do this problem
• Dec 7th 2012, 04:23 PM
topsquark
Re: Substitute x= 3 sec(theta) and simplify
Quote:

Originally Posted by mathisfun26
$\displaystyle \frac{\sqrt{x^{2}+9}}{x^{2}}$. Help write in terms of sines and cosines can anyone give me steps on how to do this problem

Well if
$\displaystyle x = 3~sec( \theta)$ then $\displaystyle x^2 = 9~sec^2( \theta )$

So
$\displaystyle \frac{\sqrt{x^2 + 9}}{x^2} = \frac{\sqrt{9~sec^2( \theta ) + 9}}{9~sec^2( \theta)}$

How do you simplify this?

-Dan
• Dec 7th 2012, 04:40 PM
skeeter
Re: Substitute x= 3 sec(theta) and simplify
Quote:

Originally Posted by mathisfun26
$\displaystyle \frac{\sqrt{x^{2}+9}}{x^{2}}$. Help write in terms of sines and cosines can anyone give me steps on how to do this problem

seems easier to use $\displaystyle x = 3\tan{\theta}$
• Dec 7th 2012, 05:58 PM
mathisfun26
Re: Substitute x= 3 sec(theta) and simplify
I'm not sure how to simplify it
• Dec 7th 2012, 06:10 PM
MarkFL
Re: Substitute x= 3 sec(theta) and simplify
Are you specifically instructed to use the given substitution? The substitution suggested by skeeter makes much more sense to use.
• Dec 7th 2012, 06:18 PM
mathisfun26
Re: Substitute x= 3 sec(theta) and simplify
Yes I was given that as the instruction
• Dec 7th 2012, 06:44 PM
MarkFL
Re: Substitute x= 3 sec(theta) and simplify
Okay, look at what topsquark posted. Try factoring the radicand, and dividing out factors common to the numerator and denominator. What do you get?
• Dec 7th 2012, 07:04 PM
mathisfun26
Re: Substitute x= 3 sec(theta) and simplify
$\displaystyle \frac{3 csc\Theta }{{}9 sec\Theta ^{^{2}}}$
im not sure this is right(Thinking)
• Dec 7th 2012, 07:22 PM
MarkFL
Re: Substitute x= 3 sec(theta) and simplify
How did you arrive at that result?
• Dec 7th 2012, 07:26 PM
mathisfun26
Re: Substitute x= 3 sec(theta) and simplify
i factored out a 9 then it became 3.. and secant +1 became csc help me please
• Dec 7th 2012, 07:41 PM
MarkFL
Re: Substitute x= 3 sec(theta) and simplify
Factoring out the 9, which the square root makes it a 3 is good:

$\displaystyle \frac{\sqrt{9\sec^2(\theta)+9}}{9\sec^2(\theta)}= \frac{\sqrt{9(\sec^2(\theta)+1)}}{9\sec^2(\theta)} =\frac{3\sqrt{\sec^2(\theta)+1}}{9\sec^2(\theta)}=$

$\displaystyle \frac{\sqrt{\sec^2(\theta)+1}}{3\sec^2(\theta)}$

However, $\displaystyle \sec^2(\theta)+1\ne\csc^2(\theta)$. You are further instructed to write in terms of sines and cosines. How can you do this with the secant function?
• Dec 7th 2012, 07:57 PM
mathisfun26
Re: Substitute x= 3 sec(theta) and simplify
secant= 1/ cosine

$\displaystyle \frac{\sqrt{{\frac{1}{cos\Theta ^{2}} +1}}}{3\frac{1}{cos\Theta ^{2}}+1}$
• Dec 7th 2012, 08:11 PM
MarkFL
Re: Substitute x= 3 sec(theta) and simplify
Where did the + 1 in the denominator come from?
• Dec 7th 2012, 08:14 PM
mathisfun26
Re: Substitute x= 3 sec(theta) and simplify
• Dec 7th 2012, 08:24 PM
MarkFL
Re: Substitute x= 3 sec(theta) and simplify
Also, I would use standard notation for the square of a trig function, i.e.:

$\displaystyle \cos^2(\theta)$

The way you have it written implies the argument (angle) is being squared.

So, we now have:

$\displaystyle \frac{\sqrt{\frac{1}{\cos^2(\theta)}+1}}{\frac{3}{ \cos^2(\theta)}}$

How can we rewrite the division as a multiplication?
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