The discussion revolves around proving the trigonometric identity 2sin²θ - 1 = sin²θ - cos²θ. Participants express uncertainty about the proof and explore the relationship between the left and right sides of the equation.
The discussion is ongoing, with participants questioning their initial assumptions and exploring different approaches to the identity. Some guidance has been offered regarding the use of trigonometric identities, but no consensus has been reached on the proof itself.
Participants note that there may be a misunderstanding regarding the values of the expressions involved, and there is an emphasis on the need to clarify existing trigonometric identities before proceeding.
The right side is not 1.Veronica_Oles said:Homework Statement
2sin2θ - 1 = sin2θ - cos2θ
Homework Equations
The Attempt at a Solution
I am unsure of how to prove these.
So far all I have is
Left side= 2sin2θ - 1
=sin2sin2-1
And I know that right side is equal to 1.
But otherwise not sure where to go from there.
I see where I went wrong now.SammyS said:The right side is not 1.
2x2 ≠ x2⋅x2 .
So certainly, 2sin2θ ≠ sin2θ ⋅sin2θ
Would I change right side to sin^2x - 1 - sin^2x?SammyS said:The right side is not 1.
2x2 ≠ x2⋅x2 .
So certainly, 2sin2θ ≠ sin2θ ⋅sin2θ
Veronica_Oles said:Homework Statement
2sin2θ - 1 = sin2θ - cos2θ
Homework Equations
The Attempt at a Solution
I am unsure of how to prove these.
So far all I have is
Left side= 2sin2θ - 1
=sin2sin2-1
And I know that right side is equal to 1.
But otherwise not sure where to go from there.
It's much simpler than this, and there is no need whatever for double-angle identies. What identities do you already know?Veronica_Oles said:Would I change right side to sin^2x - 1 - sin^2x?