Proving Trigonometric Identities

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Homework Help Overview

The discussion revolves around proving the trigonometric identity sin^6 + cos^6 = 1 - 3sin^2cos^2. Participants are exploring methods to manipulate the left-hand side to match the right-hand side of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to express sin^6 and cos^6 in terms of sine and cosine, questioning the effectiveness of their initial approaches. Others suggest using identities such as sin^2x = 1 - cos^2x to facilitate the proof.

Discussion Status

The conversation includes various attempts to prove the identity, with some participants sharing their work and seeking validation. Guidance has been offered regarding the clarity of presentation in proofs, though no consensus on a single method has been reached.

Contextual Notes

Participants are encouraged to adhere to specific mathematical conventions in their proofs, and some express uncertainty about their previous methods. There is an acknowledgment of the challenges faced in presenting mathematical arguments clearly.

acen_gr
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Homework Statement


Prove that sin6 + cos6 = 1 - 3sin2cos2



Homework Equations


(1)


The Attempt at a Solution


I tried to convert those all in terms of sine and cosine only but it didn't work.
 
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acen_gr said:

Homework Statement


Prove that sin6 + cos6 = 1 - 3sin2cos2



Homework Equations


(1)


The Attempt at a Solution


I tried to convert those all in terms of sine and cosine only but it didn't work.

Notice that [itex]\sin^6x=\left(\sin^2x\right)^3[/itex] and that [itex]\sin^2x=1-\cos^2x[/itex]

Start by trying to apply these two ideas to the LHS of the equation.
 
This is my work. Is this right?

sin6x + cos6x = 1 - 3sin2xcos2x
(sin2x)3 + cos6x = RHS
(1 - cos2x)3 + cos6 = RHS
1 - 3cos2x + 3cos4x - cos6x + cos6x = RHS
1 - 3cos2x + 3(cos2x)(cos2x) = RHS
1 - 3cos2x + 3(cos2x)(1 - sin2x) = RHS
1 - 3cos2x[1-(1 -(sin2x)] = RHS
1 - 3cos2x(sin2x) = RHS
1 - 3cos2xsin2x = 1 - 3sin2xcos2x
 
acen_gr said:
This is my work. Is this right?
Every step is mathematically equivalent to its preceding step (so you haven't broken any rules) so yes, it's right :smile:

If I'm to be a bit picky however, I'll mention that it's not necessary to write ... = RHS on every line. You should instead set out your proof as so:

LHS = ...
= ...
= ...
= RHS

You don't need to keep writing LHS either, unless you make any manipulations to both sides of the equality such as:

LHS = ...
LHS + 1 = ... + 1

Although you shouldn't do this when trying to prove LHS = RHS. If you needed to add 1 to the RHS then you can minus one to keep it balanced as so:

LHS = ...
= ... + 1 - 1

acen_gr said:
1 - 3cos2x + 3(cos2x)(cos2x) = RHS
I'd also skip this line altogether (unless it's to help you to personally keep track of things and not get confused), but if not, you should just go ahead and factorize straight away:

[tex]=1-3\cos^2x+3cos^4x[/tex]
[tex]=1-3\cos^2x(1-cos^2x)[/tex]

acen_gr said:
1 - 3cos2x(sin2x) = RHS
1 - 3cos2xsin2x = 1 - 3sin2xcos2x
Also this last line is unnecessary, but there's no harm done if you feel like doing it.
 
Thank you!:) Now I know why my teachers hate my solutions :P I'll take note of everything you've said. God bless! Hope you could help more people like me:) I'm starting to love this forum! :D
 
acen_gr said:
Thank you!:) Now I know why my teachers hate my solutions :P I'll take note of everything you've said. God bless! Hope you could help more people like me:) I'm starting to love this forum! :D

You're welcome :smile: and that's glad to hear!
 

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