My Math Forum need trig assistance

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 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.
 February 22nd, 2013, 12:18 PM #2 Senior Member     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: $\frac{3}{7}\sec^2(x)\frac{9}{4}\tan^2(x)+\frac{27} {28}\sec^2(x)=$ $\frac{27}{28}\sec^2(x)\tan^2(x)+\frac{27}{28}\sec^ 2(x)=$ $\frac{27}{28}\sec^2(x)$$\tan^2(x)+1$$$ Then using the Pythagorean identity $\tan^2(x)+1=\sec^2(x)$ we have: $\frac{27}{28}\sec^2(x)\sec^2(x)=\frac{27}{28}\sec^ 4(x)$
 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
 February 22nd, 2013, 05:56 PM #4 Senior Member     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: $\frac{27}{28}\sec^2(x)\tan^2(x)+\frac{27}{28}\sec^ 2(x)$ Think of this as: $\frac{27}{28}\sec^2(x)\cdot\tan^2(x)+\frac{27}{28} \sec^2(x)\cdot1$ Observe that both terms have $\frac{27}{28}\sec^2(x)$ as a factor, so pull that out front. $\frac{27}{28}\sec^2(x)$$\tan^2(x)+1$$$ Does that make sense so far?
 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
 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
 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) ?
 February 22nd, 2013, 06:35 PM #8 Senior Member     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: $\frac{\csc^2(x)-1}{5$$\csc(x)-1$$}$ Factor the numerator as difference of squares: $\frac{$$\csc(x)+1$$$$\csc(x)-1$$}{5$$\csc(x)-1$$}$ Divide out common factors: $\frac{\csc(x)+1}{5}$ Voila!
 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
 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)

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