document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1100909>Question 1100909</A>: <I><SPAN STYLE=\"word-break: break-all;\"> sin(x)*cos(y)=1/2[sin(x+y) + sin(x-y)] (1)<BR>\n" );
document.write( "cos(x)*sin(y)=1/2[sin(x+y) - sin(x-y)] (2)<BR>\n" );
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document.write( "It reads: verify the product-to-sum identities below using the sine sum and difference identities.\r<BR>\n" );
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document.write( "I have no clue where to begin. This is an online course and the instructor has disabled the \"show me an example\" option and the \"help me solve\" option.  I want to learn this and if you guys can teach me It would be appreciated. Thanks in advance. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #715451 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=715451\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=\"black\" face=\"times\" size=\"3\"><BR>\n" );
document.write( "The sum and difference identities for sine are these equations below<BR>\n" );
document.write( "sin(x+y) = sin(x)cos(y)+cos(x)sin(y)<BR>\n" );
document.write( "sin(x-y) = sin(x)cos(y)-cos(x)sin(y)\r<BR>\n" );
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document.write( "These identities are often found in the appendix of your trig textbook, and also in the relevant chapters of the text. Depending on your teacher, you will either have a reference sheet, they will be given in the problem, or you have to memorize these formulas. \r<BR>\n" );
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document.write( "Add the equations shown above. We do so by adding the left sides separately and then the right sides separately. On the left side we will get sin(x+y)+sin(x-y). On the right side we will have<BR>\n" );
document.write( "[ sin(x)cos(y)+cos(x)sin(y) ] + [ sin(x)cos(y)-cos(x)sin(y) ]\r<BR>\n" );
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document.write( "that messy expression above turns into<BR>\n" );
document.write( "2sin(x)cos(y)\r<BR>\n" );
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document.write( "Notice how the cos(x)sin(y) terms go away. They add up to 0 and the 0 is \"absorbed\", more or less, into the other terms. This works since x+0 = 0+x = x for any real x. <BR>\n" );
document.write( "There are two copies of the sin(x)cos(y) terms added up, so sin(x)cos(y)+sin(x)cos(y) = 2sin(x)cos(y)\r<BR>\n" );
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document.write( "So after adding those equations, we end up with sin(x+y)+sin(x-y) = 2sin(x)cos(y)\r<BR>\n" );
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document.write( "The last thing to do is multiply both sides by 1/2, then flip the equation and we end up with this:<BR>\n" );
document.write( "sin(x)*cos(y) = 1/2 [ sin(x+y)+sin(x-y) ]<BR>\n" );
document.write( "which is the first identity we wanted to prove. \r<BR>\n" );
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document.write( "It's the same kind of story with the second equation given<BR>\n" );
document.write( "cos(x)sin(y) = 1/2 [ sin(x+y)-sin(x-y) ]\r<BR>\n" );
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document.write( "Instead of adding sin(x+y) and sin(x-y), we're subtracting this time. Once again,<BR>\n" );
document.write( "sin(x+y) = sin(x)cos(y)+cos(x)sin(y)<BR>\n" );
document.write( "sin(x-y) = sin(x)cos(y)-cos(x)sin(y)\r<BR>\n" );
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document.write( "which leads to<BR>\n" );
document.write( "sin(x+y)-sin(x-y) = 2cos(x)sin(y) \r<BR>\n" );
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document.write( "when we subtract the corresponding left and right hand sides together. The sin(x)cos(y) terms cancel this time. \r<BR>\n" );
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document.write( "Multiply both sides by 1/2 and rearrange the sides and we get<BR>\n" );
document.write( "cos(x)sin(y) = 1/2 [ sin(x+y)-sin(x-y) ]<BR>\n" );
document.write( "which proves the identity<BR>\n" );
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