My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
February 22nd, 2013, 12:07 PM   #1
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

need trig assistance

Problem says the following: Factor: (3/7 sec^2(x))(9/4tan^2(x)) + 27/28sec^2(x)


I attempted to solve and finally checked answer and it states: tan^2(x) + 1

I understand that the above is a trig identity but would like to ask the forum to show/workout/explain how we came to this


Then next step says Apply trig identity and simplify: (3/7sec^2(x))(9/4tan^2(x)) + 27/28sec^2(x) = ?

Thanks for any and all help. I do apprieciate it.
mathkid is offline  
 
February 22nd, 2013, 12:18 PM   #2
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,204
Thanks: 511

Math Focus: Calculus/ODEs
Re: need trig assistance

We may write:







Then using the Pythagorean identity we have:

MarkFL is offline  
February 22nd, 2013, 05:39 PM   #3
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

Re: need trig assistance

yep thats right Mark!

I guess where I am stumped is 27/28sec^2(x)tan^2(x)+27/28sec^2(x) I know we factor here but can you show me exactly what is factored cause I don't
see how we get the tan^2(x) + 1 ? I know that sec^2(x) = tan^2(x) + 1 but I thought we factored that out?

so Mark I am seeing this: we factor out the constant and the sec^2(x) to give 27/28sec^2(x)tan^2(x) but I do not see/understand where the +1 came from?


thank Mark
mathkid is offline  
February 22nd, 2013, 05:56 PM   #4
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,204
Thanks: 511

Math Focus: Calculus/ODEs
Re: need trig assistance

After the first step, we have:



Think of this as:



Observe that both terms have as a factor, so pull that out front.



Does that make sense so far?
MarkFL is offline  
February 22nd, 2013, 06:11 PM   #5
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

Re: need trig assistance

uh yes I think so. so the times 1 when we pull everything out front we are left with the times 1 but the times 1 is essentially just 1 left there hence +1


I just have not practiced enough to have the vision to see that

how about helping me with another similair problem.


(csc^2(x) - 1)/5(csc(x)-1))

here is me trying to work this one out. the top is 1/5(-sin^2(x))(sin(x)-1) = -1/5sin^2(x)(sin(x) = ??????

I do not yet have the vision to be able to flip and skip through this stuff
mathkid is offline  
February 22nd, 2013, 06:18 PM   #6
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

Re: need trig assistance

wait...is it -1/5csc(x) ?

if I turn all into sin and cos I get this I think pull the constant out front 1/5 pull it way out to the left so I don't even have to look at the rascal right now

then I get 1/sin^2(x) -1 times sin(x) - 1

then I get (sin(x))/(sin^2(x) and -1 times -1 is still -1

now factoring the -1 out I get -1/5(1/sin(x)) and flipping that I get -1/5csc(x) is this right or wrong lol

ty
mathkid is offline  
February 22nd, 2013, 06:27 PM   #7
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

Re: need trig assistance

9/5-18/5sin^2(x)+9/4sin^4(x)




am I able to factor this down to just -9/5sin^2(x) ?
mathkid is offline  
February 22nd, 2013, 06:35 PM   #8
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,204
Thanks: 511

Math Focus: Calculus/ODEs
Re: need trig assistance

You'll have to bear with me...I have problems being sent to me by PM on two forums, moderating duties on two forums, and my sister and girlfriend by email all happening at once...

We have:



Factor the numerator as difference of squares:



Divide out common factors:



Voila!
MarkFL is offline  
February 22nd, 2013, 06:42 PM   #9
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

Re: need trig assistance

wow Mark that was so freakin simple it makes me want to puke lol now I re the teacher going over that too, it had just slipped my mind.

ps... the reason they are all pm'ing and emailing u is becuase u r the man
mathkid is offline  
February 22nd, 2013, 06:43 PM   #10
Senior Member
 
Joined: Aug 2012
From: South Carolina

Posts: 866
Thanks: 0

Re: need trig assistance

also he told us until we have the vison ... we can sub x or y for example in for cos or sin whatever to see it easier

like the one u just did x^2 -1 is (x+1)(x-1)
mathkid is offline  
Reply

  My Math Forum > High School Math Forum > Trigonometry

Tags
assistance, trig



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
need assistance nawras Algebra 3 August 18th, 2013 09:58 AM
Assistance constructing a graph iloveall Algebra 3 July 5th, 2012 10:40 PM
Matrix Assistance gretty Algebra 3 March 7th, 2010 03:11 AM
Matrice Assistance gretty Linear Algebra 1 March 5th, 2010 04:26 PM
Need Assistance Getting Reassociated with Math stinianne Algebra 4 August 31st, 2007 05:48 PM





Copyright © 2018 My Math Forum. All rights reserved.