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Trig identities

by Nelo
Tags: identities, trig
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Nelo
#1
Jul25-11, 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 1-sin^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?
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rock.freak667
#2
Jul25-11, 05:09 PM
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P: 6,202
Letter m is

[tex]\frac{1+tan^2x}{1-tan^2x} = \frac{1}{cos^2x-sin^2x}[/tex]


If you take the left side and divide both the numerator and denominator by cos2x what will you get?
Nelo
#3
Jul25-11, 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 ....

rock.freak667
#4
Jul25-11, 05:34 PM
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P: 6,202
Trig identities

Quote Quote by Nelo View Post
Divide the tan bracket by cos^2x?

you would get the original tan bracket, with a division of cos^2x ....
Sorry, wrong operation.

rewrite tanx as sin/cosx and then multiply the numerator and denominator by cos2x. It should easily work out.
Nelo
#5
Jul25-11, 05:44 PM
P: 215
Its not tan x though, Its 1 + tan^2x , cant write tan^2x as sin/cos
rock.freak667
#6
Jul25-11, 05:47 PM
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P: 6,202
Quote Quote by Nelo View Post
Its not tan x though, Its 1 + tan^2x , cant write tan^2x as sin/cos
tanx = sinx/cosx

squaring both sides

tan2x = (sinx/cosx)2 = ?
Nelo
#7
Jul25-11, 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?
SammyS
#8
Jul25-11, 05:56 PM
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Quote Quote by rock.freak667 View Post
Letter m is

[tex]\frac{1+tan^2x}{1-tan^2x} = \frac{1}{cos^2x-sin^2x}[/tex]


If you take the left side and divide both the numerator and denominator by cos2x what will you get?
How about: taking the left side and multiplying divide both the numerator and denominator by cos2x ?
Nelo
#9
Jul25-11, 08:45 PM
P: 215
does 1 - sin = cos?
rock.freak667
#10
Jul25-11, 09:40 PM
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Quote Quote by Nelo View Post
does 1 - sin = cos?
No. But 1-sin2x=cos2x
Nelo
#11
Jul26-11, 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 1-cos^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.
Nelo
#12
Jul26-11, 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 1-cosx.


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

1-cos^2x / cos

= 1 - cos.

ls = rs

Is this correct?
rock.freak667
#13
Jul26-11, 04:23 PM
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Quote Quote by Nelo View Post
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 1-cos^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.
If you take the left side, what can you replace '1' by? (something with sine and cosine in it)

Quote Quote by Nelo View Post
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 1-cosx.


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

1-cos^2x / cos

= 1 - cos.

ls = rs

Is this correct?
That is one way to do it, but you should generally use one side and prove the other.
Nelo
#14
Jul26-11, 04:30 PM
P: 215
Are you ever supposed to recipricol subtraction into addition? Or is it jus for division into multiplication?
rock.freak667
#15
Jul26-11, 04:37 PM
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P: 6,202
Quote Quote by Nelo View Post
Are you ever supposed to recipricol subtraction into addition? Or is it jus for division into multiplication?
I am not sure what you mean by that.

When you multiply the numerator and denominator by the same quantity you are essentually multiplying by 1.
Nelo
#16
Jul26-11, 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
________ - 1-cos^2x = 1-cos^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.
rock.freak667
#17
Jul26-11, 04:40 PM
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Quote Quote by Nelo View Post
Tried the left side, that only simplifies into sinx + sinx/cosx .

Any hints?
You could take this and then bring them both down the same denominator of cosx.
Nelo
#18
Jul26-11, 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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