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November 16th, 2009, 08:51 AM  #1 
Newbie Joined: Nov 2009 Posts: 5 Thanks: 0  3 Math Problems!!!
I need help with the some math problems. they are really really hard, and I want to know how to do them (not for homework, but for an upcoming test that the teacher said they would be on!). Prove the following identity: #1 (tanx + cotx)/(secx  tanx) = sec^2x(secx)(tanx) The teacher already told us that it would be an identity, but we have to be able to prove why without using other known identities. Solve the following equations: #1 12sin^2(2x) = 3sin(2x) #2 (16/81)^(sin^2(x)) + (16/81)^(1sin^2(x)) = 26/27 I am not sure how to even start this second one, and I am really nervous! Any help is greatly appreciated!!!! 
November 16th, 2009, 09:16 AM  #2 
Member Joined: Oct 2009 Posts: 39 Thanks: 0  Re: 3 Math Problems!!!
What can you know before proving these statements? I mean can you use sin^2(x)+cos^2(x)=1 and sin2x=2sinxcosx, cos2=cos^2sin^2 etc.? If yes, please just try to multiply and divide. It would lead to proving. For the last  first step please note (from rule above) that 1sin^2=cos^2. I am working on it, so I hope we could compare our results. BTW: any statement with TANKS (tanx) and SEX (secx) are difficult to prove. 
November 16th, 2009, 12:23 PM  #3  
Global Moderator Joined: Dec 2006 Posts: 21,027 Thanks: 2258  Quote:
The equation 1  2sin²(2x) = 3sin(2x) can be solved, but was 1 + 2sin²(2x) = 3sin(2x) intended? Both are quadratics in sin(2x), but the second factorizes nicely.  
November 16th, 2009, 01:16 PM  #4 
Newbie Joined: Nov 2009 Posts: 5 Thanks: 0  Re: 3 Math Problems!!!
It is intended to solve 12sin^2(2x)=3sin(2x) The teacher said that we should be able to prove the first equation (tanx+cotx)/(secxtanx)= sec^2x secxtanx, but that it would be difficult. Is there any way to simplify it? I cannot figure this one out for the life of me. Best of luck! 
November 16th, 2009, 01:40 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,027 Thanks: 2258 
The "identity" (tan x + cot x)/(sec x  tan x) = sec²x  sec x tan x must be mistyped. It isn't correct as typed. When x = pi/4, for example, the lefthand side is 2 + 2?2, but the righthand side is 2  ?2. 
November 16th, 2009, 02:00 PM  #6 
Newbie Joined: Nov 2009 Posts: 5 Thanks: 0  Re: 3 Math Problems!!!
You are right... that might be a source of my frustration... I will bring that up with him and see what he says. Thanks for noticing!

November 16th, 2009, 02:11 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 21,027 Thanks: 2258 
For real x, 1  2sin²(2x) = 3sin(2x) is satisfied if sin(2x) = (?17  3)/4. Hence x = arcsin((?17  3)/4) + 2k(pi) or pi  arcsin((?17  3)/4) + 2k(pi), where k is an integer. By inspection, the final problem has solutions x = pi/4 ± pi/12 + (k/2)pi, where k is an integer. 
November 16th, 2009, 02:22 PM  #8 
Newbie Joined: Nov 2009 Posts: 5 Thanks: 0  Re: 3 Math Problems!!!
I've been working on another problem that says: Disprove the following identities by giving counterexamples: so the first one is sin^2(x) + cos(x) = 0 I've been plugging in examples, but I am not getting any that have '0'. For example, I plugged in 2pi, 3pi/2, pi, pi/2, and they all come back greater than zero (which would make it not an identity, right?) Is it that simple? As for the help above THANK YOU!!, but I don't know how you got those. Where did you start? Especially the last one. Will you show me how to begin. I appreciate your helping me! 
November 16th, 2009, 02:38 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 21,027 Thanks: 2258 
Just one counterexample (such as x = 0) suffices. Regarding my two answers in my previous post, the first was obtained using the quadratic formula, and the second problem is solved by noticing that it is effectively an easily factorizable quadratic equation in (16/81)^(sin²x); one obtains sin²x = 1/4 or 3/4, and hence the solutions I gave. 
November 16th, 2009, 04:54 PM  #10  
Newbie Joined: Nov 2009 Posts: 5 Thanks: 0  Re: Quote:
It is pi/4 because there are two pi/2 and then it is plus or minus pi/12 because you add the pi/3 and the pi/4 but I don't understand where you get the 'add k/2 pi'.  

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