Trig identities

  • Thread starter Nelo
  • Start date
  • #1
215
0

Homework Statement



Prove this identity


m) Sin^2 x / sin^2 x + cos x^2 = tan^2 x / 1 + tan^ x

Homework Equations



http://i52.tinypic.com/105tdtk.jpg Letter (m) on the top

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?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
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?
 
  • #3
215
0
Divide the tan bracket by cos^2x?

you would get the original tan bracket, with a division of cos^2x ....
 
  • #4
rock.freak667
Homework Helper
6,223
31
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.
 
Last edited:
  • #5
215
0
Its not tan x though, Its 1 + tan^2x , cant write tan^2x as sin/cos
 
  • #6
rock.freak667
Homework Helper
6,223
31
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 = ?
 
  • #7
215
0
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
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,377
1,038
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 [STRIKE]divide[/STRIKE] both the numerator and denominator by cos2x ?
 
  • #9
215
0
does 1 - sin = cos?
 
  • #11
215
0
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.
 
  • #12
215
0
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?
 
  • #13
rock.freak667
Homework Helper
6,223
31
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)

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.
 
  • #14
215
0
Are you ever supposed to recipricol subtraction into addition? Or is it jus for division into multiplication?
 
  • #15
rock.freak667
Homework Helper
6,223
31
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.
 
  • #16
215
0
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.
 
Last edited:
  • #17
rock.freak667
Homework Helper
6,223
31
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.
 
  • #18
215
0
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.
 
  • #19
rock.freak667
Homework Helper
6,223
31
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.

It seems like you are working with both sides at the same time. It is better to just take one side and use that side to prove the other. How you wrote it is confusing.


Sin^2x
________ - 1-cos^2x = 1-cos^2x + sin^2x / cos^2x
cos^2x

From

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

best to not change sin2x as yet. If you factor out the sin2x, what are you left with?

Can you see an identity that will help to get tan2x?
 
  • #20
215
0
I already have that one solved, was just wondering if my way is valid to. And yes i know, my teaccher doesnt care though.

How about this one?

1/sinx^2x + 1/cos^2x = 1/sin^2x*cos^2x

I tried turning 1/sin^2x into cosecent^2x + secent^x = 1/sin^2x*cos^2x

Also just converted the two values at the bottom into 1-cos^2x and 1-sin^2x , however i dont understand how they are turned into multiplication on the right side?
 
  • #21
rock.freak667
Homework Helper
6,223
31
I already have that one solved, was just wondering if my way is valid to. And yes i know, my teaccher doesnt care though.

Use one side to prove the other is quicker. Reducing both sides to the same expression is equally valid when I learned it.

How about this one?

1/sinx^2x + 1/cos^2x = 1/sin^2x*cos^2x

I tried turning 1/sin^2x into cosecent^2x + secent^x = 1/sin^2x*cos^2x

Also just converted the two values at the bottom into 1-cos^2x and 1-sin^2x , however i dont understand how they are turned into multiplication on the right side?

That's over complicating it. Look at it like this, two terms are being reduced to one, so you'd need to bring them to a common denominator.

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


(I assume you are able to do this)
 
  • #22
215
0
yes, i took the cos fraction and multiplied the 1/sin^2x to give me

cos^2x + 1
______________ ____
sinx^2x*cos^2x cos^2x


the opposite side cos^2xs cancel, and you get 1/sin^2x * cos^2x.

Sorry, I am completely new to these concepts, and I have a test on them tommorow. Not much breathing room for me, kinda panicing.

Are you able to answer a question of sin/cos graphs if I asked?
 
  • #23
215
0
If i had a cos function like y=4cos1/3(x+2pi) -4

With the final points of

-360 , 0

-90, -4
180, -8
450 , -4
720, 0

How would i continue this for 2 cycles?

The period is 1080, how do i get the points for the rest of the cycle? Would i have to perform all of these translations on 2 cycles of the normal x/y graph of the cos / sin function?

ie) normally its..

0 1
90 0
180 -1
270, 0
360, 1

If i continued that to 450, would the next value be 0 ? then -1 ? then 0 ?
 
  • #24
rock.freak667
Homework Helper
6,223
31
yes, i took the cos fraction and multiplied the 1/sin^2x to give me

cos^2x + 1
______________ ____
sinx^2x*cos^2x cos^2x

okay well from

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

I will help you partially: with 1/sin2x

If you divide the common denominator (sin2xcos2x) by sin2x you will be left with cos2x and cos2x *1 (the numerator of 1/sin2x) is cos2x

Hence

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

Do the same for 1/cos2x and you will get the other '?'. Then just apply a simple trig identity.


If i had a cos function like y=4cos1/3(x+2pi) -4

With the final points of

-360 , 0

-90, -4
180, -8
450 , -4
720, 0

How would i continue this for 2 cycles?

The period is 1080, how do i get the points for the rest of the cycle? Would i have to perform all of these translations on 2 cycles of the normal x/y graph of the cos / sin function?

Every 1080 degrees then will go through one cycle. (-360 to 720 is 1080).

So for the next cycle the 'y' values will repeat for the corresponding x values. (so the change from -360 to -90 will be added to the next value and so on)
 
  • #25
11
0
As you know that sin^2 x + cos^2 x = 1 hence the Left Hand Side is Sin^2 x as the denominator is 1.

Now Prove Right Hand Side equal to sin^2 x too.

The right side is tan^2 x / 1 + tan^2 x convert everything into sin and cos as shown below;

(sin^2 x/cos^2 x)/(1 + sin^2 x/cos^2 x)

= (sin^2 x/cos^2 x)/[(sin^2 x +cos^2 x)/cos^2 x]

= (sin^2 x/cos^2 x) * cos^2 x/(sin^2 x +cos^2 x)

= Sin^2 x which is equal to Left Hand Side

Finally, if you can't prove Direct, simplify both sides into sin and cos and make them same.
 

Related Threads on Trig identities

  • Last Post
Replies
6
Views
741
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
15
Views
4K
  • Last Post
Replies
18
Views
1K
Top