Trigometric Identies Rearrangement

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Homework Help Overview

The discussion revolves around rearranging a trigonometric equation involving tangent and secant functions into a specific form. The original poster presents the equation 2tan2x + (2x-1)(2sec²2x) = 0 and seeks to transform it into the form 4x + sin4x - 2 = 0.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation using trigonometric identities and expresses uncertainty about the next steps after an initial rearrangement. Some participants suggest considering the sine double angle identity, while others question the applicability of certain transformations. There is also a discussion about the relationship between different angle formulas and their implications for the problem at hand.

Discussion Status

The discussion is active, with participants exploring various identities and transformations. Some guidance has been offered regarding the use of the sine double angle identity, but there remains a lack of consensus on the best approach to take. Multiple interpretations of the problem are being examined.

Contextual Notes

Participants are navigating the constraints of the problem, particularly regarding the manipulation of trigonometric identities and the specific forms required for the rearrangement. There is an ongoing debate about the validity of certain substitutions and transformations in the context of the original equation.

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Homework Statement



I have the term:

2tan2x + (2x-1)(2sec²2x) = 0

I need to rearrange it into the form

4x + sin4x - 2 = 0

I tried:
[tex]\frac{2sin2x}{cos2x} + \frac{2(2x-1)}{cos²2x} = 0[/tex]

i multiply each side by cos²2x

[tex]2.Sin2x.Cos2x + 4x - 2= 0[/tex]

but now where?

Thanks
 
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Look at the sine double angle identity.
 
i thought about that but you can't use it... can you?

sin2x = 2sinxcosx

you could do

2(2sinxcosx)cos2x + 4x - 2

or can you do somthing with the 2sin2xCos2x ? it's 2x not x so you can't equate it to sin2x

Not sure :/

Thanks
 
Let u = 2x

2sin(u)cos(u) = sin(2u)
 
but to equal sin4x doesn't it need to be 4sinxcosx not 2sin2xcos2x?

Thanks
 
Not at all.
Obviously 2sin(u)cos(u) = sin(2u) is true since the variable inside the formula doesn't matter. Now, replace u with 2x.
2sin(2x)cos(2x) = sin(2*(2x)) = sin(4x).
There's no need for another 2 out in front.
 
and that's ture for all double angle formulas
and similar for the half angle formulas?

Cheerz :)
 

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