# Trigonometric Identities

## Homework Statement

2sin2θ - 1 = sin2θ - cos2θ

## 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
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

2sin2θ - 1 = sin2θ - cos2θ

## 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

The right side is not 1.

2x2 ≠ x2⋅x2 .

So certainly, 2sin2θ ≠ sin2θ ⋅sin2θ
I see where I went wrong now.

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
Mentor

## Homework Statement

2sin2θ - 1 = sin2θ - cos2θ

## 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?