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

[clawkz]

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...

 

[Chestermiller]

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.

 

[clawkz]

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

 

[Chestermiller]

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.

 

[clawkz]

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]

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

 

[Infinitum]

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.

 

[clawkz]

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).

 

[Infinitum]

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.

 

[clawkz]

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 -.- .

 

[Infinitum]

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,

$$sin4x=sin2(2x)$$

 

[clawkz]

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.

 

[Infinitum]

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! As you say you understand it, why don't you write it down here? If there's any mistakes, I can point them out.

 

[clawkz]

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)?

 

[Infinitum]

sin4x--> sin2(2x)--> 2(2sinxcosx)

Umm, no

$$sin4x = sin2(2x) = sin2y$$

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.

 

[clawkz]

Umm, no

$$sin4x = sin2(2x) = sin2y$$

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)?

 

[Infinitum]

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.

 

[dimension10]

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.

 

[clawkz]

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

 

[Infinitum]

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!