document.write( "Question 253255: Could you please help me prove that sin(x+y)sin(x-y)=cos^2y-cos^2x? \r
\n" ); document.write( "\n" ); document.write( "This is what I have done so far...\r
\n" ); document.write( "\n" ); document.write( "sin(x+y)sin(x-y)can be changed to(sinxcosy+sinycosx)(sinxcosy-sinycosx) \r
\n" ); document.write( "\n" ); document.write( "Also, cos^2y-cos^2x can be changed to 1-sin^y-1-sin^2x\r
\n" ); document.write( "\n" ); document.write( "I'm not sure if I am going about this the right way, so if not, could you please point me in the right direction?
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Algebra.Com's Answer #185530 by drk(1908)\"\" \"About 
You can put this solution on YOUR website!
We have to prove:
\n" ); document.write( "(i) sin(x+y)*sin(x-y))=(cos^2y-cos^2x
\n" ); document.write( "I will simplify the left into the right.
\n" ); document.write( "identity: sin (x+y) = sinxcosy + sinycosx
\n" ); document.write( "identity: sin(x-y) = sinxcosy - sinycosx
\n" ); document.write( "from (i) we get
\n" ); document.write( "(ii) \"%28sinxcosy+%2B+sinycosx%29%2A%28sinxcosy+-+sinycosx%29\"
\n" ); document.write( "multiply and we get
\n" ); document.write( "(iii) sin^2x*cos^2y - sin^2y*cos^2x
\n" ); document.write( "identity: sin^2x = 1 - cos^2x
\n" ); document.write( "identity: sin^2y = 1 - cos^2y
\n" ); document.write( "(iii) becomes
\n" ); document.write( "(iv) (1-cos^2x)(cos^2y) - (1-cos^2y)(cos^2x)
\n" ); document.write( "simplifying (iv) we get
\n" ); document.write( "(v) cos^2y - cos^2x
\n" ); document.write( "QED
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