Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

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SUMMARY

The discussion revolves around proving the trigonometric identity sin(4x) = (4sin(x)cos(x))(1 - 2sin²(x)). Participants explore various approaches, including using the sine of a sum formula and double angle identities. The key steps involve rewriting sin(4x) as sin(2(2x)) and applying the double angle formulas for sine and cosine. Ultimately, the proof is confirmed by substituting the appropriate identities, leading to the correct expression.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sine and cosine functions.
  • Familiarity with double angle formulas for sine and cosine.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
  • Knowledge of the sine of a sum formula.
NEXT STEPS
  • Study the derivation and application of the sine double angle formula: sin(2x) = 2sin(x)cos(x).
  • Learn how to apply the sine of a sum formula: sin(a + b) = sin(a)cos(b) + cos(a)sin(b).
  • Explore the relationship between sin(4x) and sin(2(2x)) for deeper insights into trigonometric identities.
  • Practice proving other trigonometric identities using similar techniques and transformations.
USEFUL FOR

Students studying precalculus or trigonometry, educators teaching trigonometric identities, and anyone seeking to enhance their understanding of trigonometric proofs.

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Homework Statement


sin4x=(4sinxcosx)(1-2sin^2x)


Homework Equations


Trig identities.


The Attempt at a Solution


sin4x=(4sinxcosx)(1-2sin^2x)
(4sinxcosx)(cos2x)
stuck right here...
 
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clawkz said:

Homework Statement


sin4x=(4sinxcosx)(1-2sin^2x)


Homework Equations


Trig identities.

The Attempt at a Solution


sin4x=(4sinxcosx)(1-2sin^2x)
(4sinxcosx)(cos2x)
stuck right here...

Try starting with sin(4x) = sin(2x + 2x), and applying the formula for the sine of the sum of two angles.
 
Chestermiller said:
Try starting with sin(4x) = sin(2x + 2x), and applying the formula for the sine of the sum of two angles.
so it would be 2sinxcosx+2sinxcosx--->4sin2xcos2x-->is this right so far? i think I am wrong on the last part...btw I am horrible at precalc
 
clawkz said:
so it would be 2sinxcosx+2sinxcosx--->4sin2xcos2x-->is this right so far? i think I am wrong on the last part...btw I am horrible at precalc

Check the boldfaced items in the above, but the final result is correct. Just substitute the double angle formulas for sin2x and cos2x to complete the proof of the identity.
 
Chestermiller said:
Check the boldfaced items in the above, but the final result is correct. Just substitute the double angle formulas for sin2x and cos2x to complete the proof of the identity.
ok i get this-->4sin2xcos2x-->4(2sinxcosx)(1-2sin^2x)--> was I supposed to do something to the bolded 4? as the final thing is (4sinxcosx)(1-2sin^2x)
 
clawkz said:
ok i get this-->4sin2xcos2x-->4(2sinxcosx)(1-2sin^2x)--> was I supposed to do something to the bolded 4? as the final thing is (4sinxcosx)(1-2sin^2x)
never mind i get it. I was over thinking it lol
 
You can also do it from RHS to LHS

sin(4x) = (4sin\: x.cos\: x)(1-2sin^2x)

Write both the brackets in terms of sin2x and cos2x.
 
I thought I had been right but I was still lost. can you please look over my work and tell me where I went wrong. thanks
sin4x=(4sinxcosx)(1-sin^2x)--->sin4x= sin(2x+2x)--->2sinxcosx+2sinxcosx= 4sin2xcos2x-->4(2sinxcosx)(cos2x)----> I don't know how 4(2sinxcosx) is equal to (4sinxcosx).
 
clawkz said:
sin4x= sin(2x+2x)--->2sinxcosx+2sinxcosx

This part is wrong.

Whats the formula for sin(2x) in terms of x?

Once you answer that, your equation is sin2(2x), just expand it.
 
  • #10
Infinitum said:
This part is wrong.

Whats the formula for sin(2x) in terms of x?

Once you answer that, your equation is sin2(2x), just expand it.

I did that because isn't sin4x the same thing as sin2x + sin 2x? and according to my double angle formula sheet sin2x= 2sinxcosx, so this is why i did 2sinxcosx +2sinxcosx = 4sin2xcos2x . I don't really understand what you mean by "sin(2x) in terms of x" I am not the greatest in math -.- .
 
  • #11
clawkz said:
I did that because isn't sin4x the same thing as sin2x + sin 2x? and according to my double angle formula sheet sin2x= 2sinxcosx, so this is why i did 2sinxcosx +2sinxcosx = 4sin2xcos2x . I don't really understand what you mean by "sin(2x) in terms of x" I am not the greatest in math -.- .

You already have the answer! Use that bold relation!

And no, sin4x\neq sin 2x+sin2x

But,

sin4x=sin2(2x)
 
  • #12
Infinitum said:
You already have the answer! Use that bold relation!

And no, sin4x\neq sin 2x+sin2x

But,

sin4x=sin2(2x)

ok I think I understand it now. Can I ask you one last favor and please write it down as if you had just gotten the question on a quiz or something? I want to see every step by step.
so 1) Prove : sin4x=(4sinxcosx)(1-2sin^2x)
please Its late where I live and I need some sleep.
 
  • #13
clawkz said:
ok I think I understand it now. Can I ask you one last favor and please write it down as if you had just gotten the question on a quiz or something? I want to see every step by step.
so 1) Prove : sin4x=(4sinxcosx)(1-2sin^2x)
please Its late where I live and I need some sleep.

No.

By the homework guidelines, I cannot. And I preferably wouldn't. You won't enjoy math unless you do it by yourself! :smile:As you say you understand it, why don't you write it down here? If there's any mistakes, I can point them out.
 
  • #14
ok well here it goes... sin4x=(4sinxcosx)(1-2sin^2x)-->sin4x--> sin2(2x)--> 2(2sinxcosx)-->4sin2xcos2x-->...the confusing part-->4(2sinxcosx)(1-sin^2x)-->(4sinxcosx)(1-sin^2x)

on the italic part am i supposed to distribute the 2 out to 2sinxcosx so i can get the 4sin2xcos2x so that i can factor out the 4 and then have 4(sin2xcos2x) and then use double angle formula and get 4(2sinxcosx)(cos2x, which becomes1-sin^2x)?
 
Last edited:
  • #15
clawkz said:
sin4x--> sin2(2x)--> 2(2sinxcosx)

Umm, no :redface:

sin4x = sin2(2x) = sin2y

here y=2x. Now using the sin2y formula...

sin2y = 2siny.cosy

Buy you already have y=2x. So,

sin4x=2(sin2x.cos2x)

Now proceed by writing sin2x and cos2x in terms of sinx and cosx.
 
  • #16
Infinitum said:
Umm, no :redface:

sin4x = sin2(2x) = sin2y

here y=2x. Now using the sin2y formula...

sin2y = 2siny.cosy

Buy you already have y=2x. So,

sin4x=2(sin2x.cos2x)

Now proceed by writing sin2x and cos2x in terms of sinx and cosx.

ok makes more sense now, but now that i have 2(2sinxcosx)(cos2x)...does the 2 way in front get distrubuted like this--> (4sinxcosx)(cos2x) why not like this (4sin2xcos2x)(cos2x)?
 
  • #17
clawkz said:
ok makes more sense now, but now that i have 2(2sinxcosx)(cos2x)...does the 2 way in front get distrubuted like this--> (4sinxcosx)(cos2x) why not like this (4sin2xcos2x)(cos2x)?

Yes, you get (4sinx.cosx)(cos2x).

(4sin2x.cos2x)(cos2x) doesn't make sense, where did you get it from?? There wasn't a 2 along with a sin2x.cos2x to begin with.
 
  • #18
clawkz said:

Homework Statement


sin4x=(4sinxcosx)(1-2sin^2x)


Homework Equations


Trig identities.


The Attempt at a Solution


sin4x=(4sinxcosx)(1-2sin^2x)
(4sinxcosx)(cos2x)
stuck right here...


I guess this response is useless now but I would use the exponential forms of sine and cosine.
 
  • #19
thank you all who helped or tried to. I finally understand. The problem I was having was thinking 2(2sinxcosx)-->(4sin2xcos2x), but now i understand it just becomes 4sinxcosx. Really appreciate the help
 
  • #20
clawkz said:
thank you all who helped or tried to. I finally understand. The problem I was having was thinking 2(2sinxcosx)-->(4sin2xcos2x), but now i understand it just becomes 4sinxcosx. Really appreciate the help

Glad you figured it out! :biggrin:
 

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