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Trig Identity proof

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  • #1
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Show that:

[itex]4(\sin^4x+\cos^4x) \equiv \cos4x + 3[/itex].

Really stuck with this, no idea how to go ahead with it. The book gives a hint: [itex]\sin ^4 x = (\sin ^2 x)^2[/itex] and use [itex]\cos 2x = 1 - 2\sin ^2 x[/itex]

But I don't even understand the hint, where did they get
[itex]\cos 2x = 1 - 2\sin ^2 x[/itex] from?
 

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  • #2
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Show that:

[itex]4(\sin^4x+\cos^4x) \equiv \cos4x + 3[/itex].

Really stuck with this, no idea how to go ahead with it. The book gives a hint: [itex]\sin ^4 x = (\sin ^2 x)^2[/itex] and use [itex]\cos 2x = 1 - 2\sin ^2 x[/itex]

But I don't even understand the hint, where did they get
[itex]\cos 2x = 1 - 2\sin ^2 x[/itex] from?
This is one of three double-angle identities for cos(2x).

The other two are
cos(2x) = cos2(x) - sin2(x)
cos(2x) = 2cos2(x) - 1

The 2nd one above and the one you're asking about can be obtained from the first one I showed by using the identity sin2(x) + cos2(x) = 1.
 
  • #3
CAF123
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But I don't even understand the hint, where did they get
cos2x=1−2sin2x from?
[tex] cos (x + x) = cos^2x - sin^2x = (1 - sin^2x) -sin^2x = 1 - 2sin^2x, [/tex] using the identity [itex] sin^2x +cos^2x =1 [/itex]
 
  • #4
CAF123
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To start you off, write the given equation as, [tex] 4((\sin^2x)^2 + (\cos^2x)^2), [/tex] and use the double angle identities to get this solely in terms of [itex] \cos.[/itex]
 
  • #5
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You can start by using this:
a2+b2=(a+b)2-2ab
This would mean that:
[tex]4(\sin^4x+\cos^4x)=4((sin^2x+cos^2x)^2-2sin^2xcos^2x)[/tex]
Now rest should be easy to solve.
 
  • #6
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To start you off, write the given equation as, [tex] 4((\sin^2x)^2 + (\cos^2x)^2), [/tex] and use the double angle identities to get this solely in terms of [itex] \cos.[/itex]
working :

[itex]4((sin^2x)^2 + (cos^4x) [/itex]
[itex]4((1-cos^2x)^2 + cos^4x))[/itex]
[itex]4(1-2cos^2x + cos^4x + cos^4x) [/itex]
[itex]4(1-2cos^2x + 2cos^4x) [/itex]

Pretty much stuck here.
You can start by using this:
a2+b2=(a+b)2-2ab
This would mean that:
[tex]4(\sin^4x+\cos^4x)=4((sin^2x+cos^2x)^2-2sin^2xcos^2x)[/tex]
Now rest should be easy to solve.
continuing from your working: [itex]4(1 - 2sin^2xcos^2x)[/itex], but I don't really know any identities to go from here...
 
  • #7
Bacle2
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How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?
 
  • #8
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How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?
He doesn't have [itex] cos(4x) [/itex], he has [itex] cos^{4}(x) [/itex], or [itex] (cos(x))^{4} [/itex]
 
  • #9
ehild
Homework Helper
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working :

[itex]4((sin^2x)^2 + (cos^2x)^2) [/itex]
Use the following double-angle identities:
sin2(x)=0.5(1-cos(2x))
cos2(x)=0.5(1+cos(2x))

ehild
 
Last edited:
  • #10
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How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?
He doesn't have [itex] cos(4x) [/itex], he has [itex] cos^{4}(x) [/itex], or [itex] (cos(x))^{4} [/itex]
Bacle2 explicitly was referring to the right side of the original identity, which has a cos(4x) term in it.

See below.

[itex]4(\sin^4x+\cos^4x) \equiv \cos4x + 3[/itex].
 
  • #11
Bacle2
Science Advisor
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He doesn't have [itex] cos(4x) [/itex], he has [itex] cos^{4}(x) [/itex], or [itex] (cos(x))^{4} [/itex]
Look at the RH side of the original post, it has a Cos(4x)+3.


Edit: Sorry, I did not see Mark44's post, and now I cannot see the 'delete' option.
 
Last edited:
  • #12
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continuing from your working: [itex]4(1 - 2sin^2xcos^2x)[/itex], but I don't really know any identities to go from here...
Try using these identities:
i)sin(2x)=2sin(x)cos(x)
ii)cos(2x)=1-2sin2x

If you square both the sides of i identity, you may see a way through.
 
  • #13
CAF123
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Use what ehild suggested if attempting the way I started (which is just one way to tackle the problem - there are many others).
We have, as suggested in my last post,
[itex] 4((\sin^2x)^2 + (\cos^2x)^2) [/itex]

Now, use the double angle identities that ehild suggested, specifically [itex] \sin^2x = -\frac{1}{2}(\cos2x - 1) [/itex] and [itex] \cos^2x = \frac{1}{2}(cos2x +1) [/itex]

Substitute these into the eqn, expand the brackets and simplify...
 

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