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

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Homework Help Overview

The discussion revolves around proving the trigonometric identity sin4x=(4sinxcosx)(1-2sin^2x). Participants are exploring the application of trigonometric identities and the sine addition formulas in the context of this identity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to manipulate the identity using known trigonometric identities, including the sine of a sum and double angle formulas. Questions arise regarding the correctness of their transformations and the relationships between different sine and cosine expressions.

Discussion Status

Some participants express confusion about specific steps in their reasoning, while others provide guidance on how to approach the problem. There is a recognition of the need to clarify certain transformations and the application of identities, but no explicit consensus is reached on a single method.

Contextual Notes

Participants mention feeling stuck at various points and question their understanding of the relationships between sine and cosine functions. There is an acknowledgment of homework guidelines that discourage providing complete solutions, emphasizing the learning process instead.

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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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