Trigonometry Identities.

  • Thread starter pavadrin
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  • #1
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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
 

Answers and Replies

  • #2
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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!
 
  • #3
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oh, my mistake i forgot the [itex]\theta[/itex] sign after the trigonometric functions. i am not sure on how top factorise, sorry :(
 
  • #4
HallsofIvy
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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?
 
  • #5
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can [itex]x^2 + y^2[/itex] be factorised? ive 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
 
  • #6
BobG
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pavadrin said:
can [itex]x^2 + y^2[/itex] be factorised? ive 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].
 
  • #7
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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.
 
  • #8
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okay thanks for the help people, sorry to be a little slow
 
  • #9
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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
VietDao29
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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
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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
HallsofIvy
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Unfortunately, I have had students declare that
[tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!
 
  • #13
Office_Shredder
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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
VietDao29
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HallsofIvy said:
Unfortunately, I have had students declare that
[tex]\frac{sin x}{sin y}= \frac{x}{y}[/tex]!
... :rolleyes: *no comments* :)
 
  • #15
BobG
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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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