document.write( "Question 188285: Verify that the equation is an identity.
\n" ); document.write( " $\"sin2x%2F2sinx+=+cos%5E2+x%2F2+-+sin%5E2+x%2F2\"$
\n" ); document.write( "I tried solving on the right side
\n" ); document.write( " = [(1 + cos x) / 2] - [(1 - cos x) / 2]
\n" ); document.write( " = [1 + cosx - 1 + cos x]/2
\n" ); document.write( " = 2cosx/2
\n" ); document.write( " = cosx
\n" ); document.write( "I have no clue how to get sin2x over 2sinx from this and I don't know what I did wrong. \n" ); document.write( "

Algebra.Com's Answer #141139 by Alan3354(69443) $\"\"$ $\"About$

You can put this solution on YOUR website!
Verify that the equation is an identity.
\n" ); document.write( " $\"sin2x%2F2sinx+=+cos%5E2+x%2F2+-+sin%5E2+x%2F2\"$
\n" ); document.write( "I tried solving on the right side
\n" ); document.write( " = [(1 + cos x) / 2] - [(1 - cos x) / 2]
\n" ); document.write( " = [1 + cosx - 1 + cos x]/2
\n" ); document.write( " = 2cosx/2
\n" ); document.write( " = cosx
\n" ); document.write( "I have no clue how to get sin2x over 2sinx from this and I don't know what I did wrong.
\n" ); document.write( "--------------------
\n" ); document.write( "Picking up where you left off:
\n" ); document.write( "sin(2x) = 2sin(x)cos(x)
\n" ); document.write( "Dividing that by 2sin(x) --> cos(x)
\n" ); document.write( " \n" ); document.write( "

\n" );