SOLUTION: Verify that the equation is an identity. {{{sin2x/2sinx = cos^2 x/2 - sin^2 x/2}}} I tried solving on the right side = [(1 + cos x) / 2] - [(1 - cos x) / 2] = [1 + cosx - 1 +

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Question 188285: Verify that the equation is an identity.
AMP Parsing Error of [sin2x/2sinx = cos^2 x/2 - sin^2 x/2]: Invalid function '/2-sin^2x/2': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. .
I tried solving on the right side
= [(1 + cos x) / 2] - [(1 - cos x) / 2]
= [1 + cosx - 1 + cos x]/2
= 2cosx/2
= cosx
I have no clue how to get sin2x over 2sinx from this and I don't know what I did wrong.

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Verify that the equation is an identity.
AMP Parsing Error of [sin2x/2sinx = cos^2 x/2 - sin^2 x/2]: Invalid function '/2-sin^2x/2': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. .
I tried solving on the right side
= [(1 + cos x) / 2] - [(1 - cos x) / 2]
= [1 + cosx - 1 + cos x]/2
= 2cosx/2
= cosx
I have no clue how to get sin2x over 2sinx from this and I don't know what I did wrong.
--------------------
Picking up where you left off:
sin(2x) = 2sin(x)cos(x)
Dividing that by 2sin(x) --> cos(x)

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