How can I prove the trigonometry identity \sin^4 - \cos^4 = 1 - 2\cos^2?

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Homework Help Overview

The discussion revolves around proving the trigonometric identity \(\sin^4 - \cos^4 = 1 - 2\cos^2\). Participants explore various approaches to manipulate both sides of the equation, particularly focusing on the left-hand side (LHS) and right-hand side (RHS) of the identity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to manipulate the LHS but expresses uncertainty about handling the fourth powers. Some participants suggest using factorization techniques related to the difference of squares. Others question the need for factorization and propose alternative methods involving trigonometric identities.

Discussion Status

Participants are actively engaging with the problem, offering hints and discussing various approaches. Some guidance has been provided regarding factorization and the application of known trigonometric identities, although no consensus on a complete solution has been reached.

Contextual Notes

There are mentions of missing angle notation in the trigonometric functions, which may affect the clarity of the identity being discussed. Additionally, some participants express confusion about basic factorization concepts, indicating a potential gap in foundational knowledge.

pavadrin
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i need to prove the following identity

[tex]\sin^4 - \cos^4 = 1 - 2\cos^2[/tex]

i have tried approaching this identity by solving the RHS (right hand side) of the equation but this lead no where. however i am unsure on how to maniplulate the LHS of this equation because of the power of 4. would somebody please be able to give me a slight hint in which direction to head to prove this?
many thanks,
Pavadrin
 
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Do you know how to factorize [tex]a^2-b^2[/tex]? How can you use this to help you factorize [tex]a^4-b^4[/tex]?

Also, the identity should have a variable (i.e. instead of [tex]sin^4[/tex], it should be [tex]sin^4x[/tex]).

All the best!
 
oh, my mistake i forgot the [itex]\theta[/itex] sign after the trigonometric functions. i am not sure on how top factorise, sorry :(
 
Long, long, long, before you learn about trigonometry, you should have learned that a2- b2= (a- b)(a+ b). And if we replace a by x2 and b by y2, we get that
x4- y4= (x2- y2)(x2+ y2). And what is x2+ y2 in this case?
 
can [itex]x^2 + y^2[/itex] be factorised? I've tried [itex](x + y)(x + y)[/itex], [itex](x - y)(x - y)[/itex], but neither have worked. am i completely missing the point here? its been a while since i have had to factorise. thanks
 
pavadrin said:
can [itex]x^2 + y^2[/itex] be factorised? I've tried [itex](x + y)(x + y)[/itex], [itex](x - y)(x - y)[/itex], but neither have worked. am i completely missing the point here? its been a while since i have had to factorise. thanks
In this case you don't have to, since [itex]x^2=sin^2\theta[/itex] and [itex]y^2=cos^2\theta[/itex].
 
we don't need to factorize. Just use:

[tex]sin^{4} \theta = sin^{2} \theta \cdot sin^{2} \theta[/tex]

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

[tex](1-cos^{2} \theta) \cdot (1-cos^{2} \theta) = 1 - 2\cdot cos^{2} \theta + cos^{4} \theta[/tex]

Now substract [itex]cos^{4} \theta[/itex] and you got your proof.
 
okay thanks for the help people, sorry to be a little slow
 
In addition to what these people said: you should try to remember identities that will serve as shortcuts when proving identities.

like: Sin^2 + Cos^2=1 <= this may be the most important one

Sin^2= 1 - Cos^2

Cos^2= 1 - Sin^2
 
  • #10
hmm? said:
like: Sin^2 + Cos^2=1 <= this may be the most important one

Sin^2= 1 - Cos^2

Cos^2= 1 - Sin^2
You're missing an angle here.
It should read sin2x, not just sin2 :)
 
  • #11
VietDao29 said:
You're missing an angle here.
It should read sin2x, not just sin2 :)

You're absolutely right--all those -1 from omitted thetas and xs from test questions still haven't sunk in :/...haha.
 
  • #12
Unfortunately, I have had students declare that
[tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!
 
  • #13
HallsofIvy said:
Unfortunately, I have had students declare that
[tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!

There aren't any parentheses, so it could just be that s, i, and n are all variables
 
  • #14
HallsofIvy said:
Unfortunately, I have had students declare that
[tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!
... :rolleyes: *no comments* :)
 
  • #15
HallsofIvy said:
Unfortunately, I have had students declare that
[tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!
Well, I used to think [tex]\frac{sin x}{x} = sin[/tex] but then I found out it's worse than that. It's not just any sin, it's an X-rated video filmed in someone's kitchen sinc.
 

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