Algebraic Proofs and Verifying identities

[abbarajum]

Hey all! I am having some trouble with a certain problem on my homework. I would like some guidance. I have to prove one side of the equation is equal to the other, as you may know, as this is an algebraic proof. This in itself isn't too hard. The hard part is just this one particular problem. I may be going about this the wrong way... but the problem seems really complex.

Statement to prove: sin⁴3x - cos⁴3x = 1 - 2cos²3x

I first tried starting with the right hand side.

1 - 2cos²3x = - (2cos²3x - 1)

- (2cos²3x - 1) = - cos2(3x)

I don't know if I am allowed to call that - cos6x, but I did. And I looked up the formula for cos6x, which is:

32cos⁶x - 48 cos⁴x + 18 cos²x - 1

Therefore, - cos6x = 1- 32cos⁶x + 48 cos⁴x - 18 cos²x

I even tried expanding cos3x first, then squaring it. I got the same exact terms, and wasn't really sure what to do with them:

1 - 2cos²3x = 1 - 2 (4cos³x - 3cosx)2

1 - 2 (4cos³x - 3cosx)2 = 1 - 2 (16cos⁶x - 24cos⁴x + 9cos²x)

1 - 2 (16cos6x - 24cos⁴x + 9cos²x) = 1 - 32cos⁶x + 48cos⁴x - 18cos²x

So I got stuck there, and decided to go with the left hand side.

Starting with the LHS:

sin⁴3x - cos⁴3x = (3sinx - 4sin³x)⁴ - (4cos3 - 3cosx)⁴

So, I tried to use that formula for when you have (a - b)⁴, and here's what I ended up getting:

(81sin⁴x - 432sin⁶x + 864sin⁸x - 768sin¹⁰x + 256sin¹²x) - (256cos¹²x - 768cos¹⁰x + 864cos⁸x - 432cos⁶x +81cos⁴x)

Then, expanding further and rearranging (from smallest degrees to largest degrees):

81sin⁴x - 81cos⁴x - 432sin⁶x + 432cos⁶x + 864sin⁸x - 864cos⁸x - 768sin¹⁰x + 768cos¹⁰x + 256sin¹²x - 256cos¹²x

So I am pretty much stuck here. I mean, I see a pattern when I begin with the LHS, but I'm not sure if I am on the right track. Am I missing something, or am I overthinking? Am I way off? Or did I just make a little mistake? Or can I go further?

I will be happy to clarify something - this can be weird to type out.

Thanks!

 

[I like Serena]

Hey all! I am having some trouble with a certain problem on my homework. I would like some guidance. I have to prove one side of the equation is equal to the other, as you may know, as this is an algebraic proof. This in itself isn't too hard. The hard part is just this one particular problem. I may be going about this the wrong way... but the problem seems really complex.

Statement to prove: sin⁴3x - cos4³x = 1 - 2cos²3x

I first tried starting with the right hand side.

1 - 2cos²3x = - (2cos²3x - 1)

- (2cos²3x - 1) = - cos2(3x)

I don't know if I am allowed to call that - cos6x, but I did. And I looked up the formula for cos6x, which is:

32cos⁶x - 48 cos⁴x + 18 cos²x - 1

Therefore, - cos6x = 1- 32cos⁶x + 48 cos⁴x - 18 cos²x

I even tried expanding cos3x first, then squaring it. I got the same exact terms, and wasn't really sure what to do with them:

1 - 2cos²3x = 1 - 2 (4cos³x - 3cosx)2

1 - 2 (4cos³x - 3cosx)2 = 1 - 2 (16cos⁶x - 24cos⁴x + 9cos²x)

1 - 2 (16cos6x - 24cos⁴x + 9cos²x) = 1 - 32cos⁶x + 48cos⁴x - 18cos²x

So I got stuck there, and decided to go with the left hand side.

Starting with the LHS:

sin⁴3x - cos⁴3x = (3sinx - 4sin³x)⁴ - (4cos3 - 3cosx)⁴

So, I tried to use that formula for when you have (a - b)⁴, and here's what I ended up getting:

(81sin⁴x - 432sin⁶x + 864sin⁸x - 768sin¹⁰x + 256sin¹²x) - (256cos¹²x - 768cos¹⁰x + 864cos⁸x - 432cos⁶x +81cos⁴x)

Then, expanding further and rearranging (from smallest degrees to largest degrees):

81sin⁴x - 81cos⁴x - 432sin⁶x + 432cos⁶x + 864sin⁸x - 864cos⁸x - 768sin¹⁰x + 768cos¹⁰x + 256sin¹²x - 256cos¹²x

So I am pretty much stuck here. I mean, I see a pattern when I begin with the LHS, but I'm not sure if I am on the right track. Am I missing something, or am I overthinking? Am I way off? Or did I just make a little mistake? Or can I go further?

I will be happy to clarify something - this can be weird to type out.

Thanks!

Hi abbarajum! Welcome to MHB! :)

I guess you can do it like that, but as you can see it becomes long and complex.

How about substituting:
$$\sin^2(3x) = 1 - \cos^2(3x)$$
Then you have $\cos(3x)$ everywhere and no other trig functions.
They should cancel then.

 

[Euge]

Hey all! I am having some trouble with a certain problem on my homework. I would like some guidance. I have to prove one side of the equation is equal to the other, as you may know, as this is an algebraic proof. This in itself isn't too hard. The hard part is just this one particular problem. I may be going about this the wrong way... but the problem seems really complex.

Statement to prove: sin⁴3x - cos4³x = 1 - 2cos²3x

I first tried starting with the right hand side.

1 - 2cos²3x = - (2cos²3x - 1)

- (2cos²3x - 1) = - cos2(3x)

I don't know if I am allowed to call that - cos6x, but I did. And I looked up the formula for cos6x, which is:

32cos⁶x - 48 cos⁴x + 18 cos²x - 1

Therefore, - cos6x = 1- 32cos⁶x + 48 cos⁴x - 18 cos²x

I even tried expanding cos3x first, then squaring it. I got the same exact terms, and wasn't really sure what to do with them:

1 - 2cos²3x = 1 - 2 (4cos³x - 3cosx)2

1 - 2 (4cos³x - 3cosx)2 = 1 - 2 (16cos⁶x - 24cos⁴x + 9cos²x)

1 - 2 (16cos6x - 24cos⁴x + 9cos²x) = 1 - 32cos⁶x + 48cos⁴x - 18cos²x

So I got stuck there, and decided to go with the left hand side.

Starting with the LHS:

sin⁴3x - cos⁴3x = (3sinx - 4sin³x)⁴ - (4cos3 - 3cosx)⁴

So, I tried to use that formula for when you have (a - b)⁴, and here's what I ended up getting:

(81sin⁴x - 432sin⁶x + 864sin⁸x - 768sin¹⁰x + 256sin¹²x) - (256cos¹²x - 768cos¹⁰x + 864cos⁸x - 432cos⁶x +81cos⁴x)

Then, expanding further and rearranging (from smallest degrees to largest degrees):

81sin⁴x - 81cos⁴x - 432sin⁶x + 432cos⁶x + 864sin⁸x - 864cos⁸x - 768sin¹⁰x + 768cos¹⁰x + 256sin¹²x - 256cos¹²x

So I am pretty much stuck here. I mean, I see a pattern when I begin with the LHS, but I'm not sure if I am on the right track. Am I missing something, or am I overthinking? Am I way off? Or did I just make a little mistake? Or can I go further?

I will be happy to clarify something - this can be weird to type out.

Thanks!

Hi abbarajum,

If you consider the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$, you can write

$$\sin^4 3x - \cos^4 3x = (\sin^2 3x - \cos^2 3x)(\sin^2 3x + \cos^2 3x).$$

Use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to simplify the RHS of the above equation to $1 - 2\cos^2 3x$.

 

[abbarajum]

Hi abbarajum! Welcome to MHB! :)

I guess you can do it like that, but as you can see it becomes long and complex.

How about substituting:
$$\sin^2(3x) = 1 - \cos^2(3x)$$
Then you have $\cos(3x)$ everywhere and no other trig functions.
They should cancel then.

Hi, I like Serena!

Thanks for your response.

Where would you substitute sin23x? Would you substitute it on the right hand side, so that:

1 - 2cos²3x = 2sin²3x

Thanks again.

(Sorry if this question seems a little silly.)

- - - Updated - - -

Hi abbarajum,

If you consider the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$, you can write

$$\sin^4 3x - \cos^4 3x = (\sin^2 3x - \cos^2 3x)(\sin^2 3x + \cos^2 3x).$$

Use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to simplify the RHS of the above equation to $1 - 2\cos^2 3x$.

Hi, Euge.

Ugh, that makes it so much easier! Why didn't I think of that earlier?

Thanks so much! :)

 

[I like Serena]

Hi, I like Serena!

Thanks for your response.

Where would you substitute sin23x? Would you substitute it on the right hand side, so that:

1 - 2cos²3x = 2sin²3x

Thanks again.

(Sorry if this question seems a little silly.)

I suggest doing it on the LHS:
$$\sin^4(3x) - \cos^4(3x) = (\sin^2(3x))^2 - \cos^4(3x)
= (1 - \cos^2(3x))^2 - \cos^4(3x)$$
Or you can do it as Euge suggested.

 

[abbarajum]

Okay, I got it now - way easier than I originally thought it to be.

Thanks for both of your responses. I really appreciate them.

Have a great day!