Substitute x= 3 sec(theta) and simplify

**mathisfun26** · December 7th 2012, 04:16 PM

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

**topsquark** · December 7th 2012, 04:23 PM

Originally Posted by mathisfun26

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

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

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

How do you simplify this?

-Dan

**skeeter** · December 7th 2012, 04:40 PM

Originally Posted by mathisfun26

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

seems easier to use $x = 3\tan{\theta}$

**mathisfun26** · December 7th 2012, 05:58 PM

I'm not sure how to simplify it

**MarkFL** · December 7th 2012, 06:10 PM

Are you specifically instructed to use the given substitution? The substitution suggested by skeeter makes much more sense to use.

**mathisfun26** · December 7th 2012, 06:18 PM

Yes I was given that as the instruction

**MarkFL** · December 7th 2012, 06:44 PM

Okay, look at what topsquark posted. Try factoring the radicand, and dividing out factors common to the numerator and denominator. What do you get?

**mathisfun26** · December 7th 2012, 07:04 PM

$\frac{3 csc\Theta }{{}9 sec\Theta ^{^{2}}}$
im not sure this is right

**MarkFL** · December 7th 2012, 07:22 PM

How did you arrive at that result?

**mathisfun26** · December 7th 2012, 07:26 PM

i factored out a 9 then it became 3.. and secant +1 became csc help me please

**MarkFL** · December 7th 2012, 07:41 PM

Factoring out the 9, which the square root makes it a 3 is good:

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

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

However, $\sec^2(\theta)+1\ne\csc^2(\theta)$ . You are further instructed to write in terms of sines and cosines. How can you do this with the secant function?

**mathisfun26** · December 7th 2012, 07:57 PM

secant= 1/ cosine

$\frac{\sqrt{{\frac{1}{cos\Theta ^{2}} +1}}}{3\frac{1}{cos\Theta ^{2}}+1}$

**MarkFL** · December 7th 2012, 08:11 PM

Where did the + 1 in the denominator come from?

**mathisfun26** · December 7th 2012, 08:14 PM

my bad my mistake..

**MarkFL** · December 7th 2012, 08:24 PM

Also, I would use standard notation for the square of a trig function, i.e.:

$\cos^2(\theta)$

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

So, we now have:

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

How can we rewrite the division as a multiplication?

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Re: Substitute x= 3 sec(theta) and simplify

Re: Substitute x= 3 sec(theta) and simplify

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