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Trigonometric Identities

  • #1

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.
 

Answers and Replies

  • #2
SammyS
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Science Advisor
Homework Helper
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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
 
  • #3
The right side is not 1.

2x2 ≠ x2⋅x2 .

So certainly, 2sin2θ ≠ sin2θ ⋅sin2θ
I see where I went wrong now.
 
  • #4
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?
 
  • #5
33,264
4,964

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?
 

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