What is the proof for 2sin2θ - 1 = sin2θ - cos2θ?

[Veronica_Oles]

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.

 

[SammyS]

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.

The right side is not 1.

2x² ≠ x²⋅x² .

So certainly, 2sin²θ ≠ sin²θ ⋅sin²θ

 

[Veronica_Oles]

The right side is not 1.

2x² ≠ x²⋅x² .

So certainly, 2sin²θ ≠ sin²θ ⋅sin²θ

I see where I went wrong now.

 

[Veronica_Oles]

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?

 

[Mark44]

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.

Would I change right side to sin^2x - 1 - sin^2x?

It's much simpler than this, and there is no need whatever for double-angle identies. What identities do you already know?