
#1
Jul2511, 04:23 PM

P: 215

1. The problem statement, all variables and given/known data
Prove this identity m) Sin^2 x / sin^2 x + cos x^2 = tan^2 x / 1 + tan^ x 2. Relevant equations http://i52.tinypic.com/105tdtk.jpg Letter (m) on the top 3. The attempt at a solution I dont know to much about trig identities, he barely taught anything. But apparently 1 is = to cos/cos and plenty of other things. I tried to solve the right side, the first thing i did was turn the 1 into sin^ x + cos ^ x. Then i turned the denominator into 1sin^x  1 + cos^x by subbing the values of cos^x , since cos^x = 1  sin^2 . I know that tan^x = sin^x / cos^x , but the things that I am subbing in keep cancelling eachother out, does anyone know how to solve this? 



#2
Jul2511, 05:09 PM

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P: 6,213

Letter m is
[tex]\frac{1+tan^2x}{1tan^2x} = \frac{1}{cos^2xsin^2x}[/tex] If you take the left side and divide both the numerator and denominator by cos^{2}x what will you get? 



#3
Jul2511, 05:18 PM

P: 215

Divide the tan bracket by cos^2x?
you would get the original tan bracket, with a division of cos^2x .... 



#4
Jul2511, 05:34 PM

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P: 6,213

Trig identitiesrewrite tanx as sin/cosx and then multiply the numerator and denominator by cos^{2}x. It should easily work out. 



#5
Jul2511, 05:44 PM

P: 215

Its not tan x though, Its 1 + tan^2x , cant write tan^2x as sin/cos




#6
Jul2511, 05:47 PM

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P: 6,213

squaring both sides tan^{2}x = (sinx/cosx)^{2} = ? 



#7
Jul2511, 05:53 PM

P: 215

tru, i dont get what to do when theres a 1. it seem as tho sometimes people change the 1 into sinx^2 + cosx^2, sometimes they keep the one, honostly I dont even get how to do it. Am i always looking at both side of the equation? or just focusing on one? If i pick leftside do i ignore right side to the extent of just trying to solve for it? wth do i do with the ones?




#8
Jul2511, 05:56 PM

Emeritus
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#9
Jul2511, 08:45 PM

P: 215

does 1  sin = cos?




#10
Jul2511, 09:40 PM

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P: 6,213





#11
Jul2611, 03:45 PM

P: 215

f) 2sin^2x  1 = sin^2 x  cos^2 x
I dont know how to solve this. I tried taking the right side and simplyiing it to 1cos^2x 1  sin^2x ... then didnt know where to go with it, And I have no idea where to even begin with the left side. 



#12
Jul2611, 03:57 PM

P: 215

Can anyone tell me if i did this question right?
It goes :: 1 / cosx  cosx = sinx*tanx I took the right side and simplified it to 1cosx. Then i took the left side and multiplied the lone cos value with cos to give me 1/cos  cos^2x / cos 1 / cos  cos^2x / cos 1cos^2x / cos = 1  cos. ls = rs Is this correct? 



#13
Jul2611, 04:23 PM

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#14
Jul2611, 04:30 PM

P: 215

Are you ever supposed to recipricol subtraction into addition? Or is it jus for division into multiplication?




#15
Jul2611, 04:37 PM

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When you multiply the numerator and denominator by the same quantity you are essentually multiplying by 1. 



#16
Jul2611, 04:37 PM

P: 215

Is this one right?
I solved it two ways, but im uncertain if one of these ways is right or not. f) Tan^2x  sin^2x = sin^2x*tan^2x First: Sin^2x ________  1cos^2x = 1cos^2x + sin^2x / cos^2x cos^2x ^ factor out the negetive from the cos^2x Sin^2x  1 rs cancel out the two cos^2x, and factor out the negitive giving 1 sin^2x in the end. Ls = sin^2x 1 , rs = 1cos^2x Ls = RS? The other way is the way the teacher did it, just solving one side. etc. 



#17
Jul2611, 04:40 PM

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#18
Jul2611, 04:59 PM

P: 215

Like... when you have division, you take the term on the right and flip it,, do you do that for subtraction into addition as wel?
I edited my last post for a different question, please take a look @ it. 


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