Prove the following identity algebraically

  • #1
445
0

Homework Statement


PhysicsForumcommarch21sttrig.png


Must be proven algebraically, duh!

Homework Equations


trig identities


The Attempt at a Solution



PhysicsForumcommarch21sttrigpart2.png


PhysicsForumcommarch21sttrigpart3.png


I'm at a loss as what to do next. Any help would be appreciated.
 
Last edited by a moderator:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
try using [itex]sin^2x+cos^2x=1[/itex] and substitute for cos2x
 
  • #3
symbolipoint
Homework Helper
Education Advisor
Gold Member
6,062
1,131
Another approach, from the lefthand side of the original given equation, do you see that
(csc2x-cot2x)2=the expression originally on the leftside.
 
Last edited:
  • #4
symbolipoint
Homework Helper
Education Advisor
Gold Member
6,062
1,131
Yes, again, do as in #3. Then do the substitutions for the meaning of csc and for cot, simplify using algebra steps, and use the identity suggested in post #2. This becomes very uncomplicated.
 
  • #5
445
0
[tex] \frac{2 - cos^2x + 2cos^4x}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

How am I going to get rid of the two on the L.H.S.? I can't just directly cancel it, because of the [tex] - cos^2x [/tex] right?

I'm at a loss, even with all this help. So I'm going to show you my next step and maybe you people can tell me what I'm doing wrong.

[tex] \frac{2 - [cos^2x ( 1 + 2cos^2x )]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 + cos^2x - (2sin^2xcos^2x)}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

I do this next: I cancel the two coefficients in the denominator and in the bracketed term, then I subtract the exponent from the [tex]sin^2x[/tex] in the brackets from the denominator.

I'm left with this:

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

which simplifies to:

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

Where did I go wrong?
 
  • #6
445
0
Let me restart from here:

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2 - 2sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [3cos^2x - 2sin^2xcos^2x]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

In the previous step I subtracted the exponents for sin in the denominator from the numerator. Is this a legal maneuver?

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

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

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

Is this correct? Thanks again for the help guys.
 
Last edited:
  • #7
445
0
Let me restart from here:

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2 - 2sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [3cos^2x - 2sin^2xcos^2x]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

In the previous step I subtracted the exponents for sin in the denominator from the numerator. Is this a legal maneuver?

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

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

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

Is this correct? Thanks again for the help guys.

I made in error in writing down my work, it should read:

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

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

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

[tex]\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]
 
  • #8
81
0
I would like to point out an error you made very early on.

You started with:
[tex]csc^{4}x - 2*csc^{2}x*cot^{2}x + cot^{4}x = 1[/tex]

And somehow got to:
[tex]\frac{1}{sin^{4}x} - \frac{cos^{2}x}{2*sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1[/tex]

How did your coefficient of the second term flip from the numerator to the denominator? It should read:
[tex]\frac{1}{sin^{4}x} - \frac{2*cos^{2}x}{sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1[/tex]

From there it is pretty straightforward. Add the terms together, factor the top, then use the Pythagorean trig identities to reduce it to 1.
 
  • #9
445
0
Thanks for the constructive criticism.
 

Related Threads on Prove the following identity algebraically

  • Last Post
Replies
4
Views
1K
Replies
6
Views
903
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
932
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
16
Views
2K
Top