The equation 2sin²θ - 1 = sin²θ - cos²θ can be proven using fundamental trigonometric identities. The left side simplifies to sin²(2θ) using the double angle identity for sine, while the right side can be rewritten as sin²θ - (1 - sin²θ), which simplifies to 2sin²θ - 1. Therefore, both sides are equal, confirming the identity.
PREREQUISITESStudents studying trigonometry, mathematics educators, and anyone looking to strengthen their understanding of trigonometric identities and proofs.
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?