document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=21632>Question 21632</A>: <I><SPAN STYLE=\"word-break: break-all;\"> prove the following equation is an identity. show your work<BR>\n" );
document.write( "sin5x+sin3x=4sin2xcos2xcosx\r<BR>\n" );
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document.write( "first i did the regular formula sinA+sinB=2sin(a+b/2)cos(a-b/2) so it came out like this : 2sin4xcosx and now i'm stuck and i dont know where to go from that can anyone help? thanks\r<BR>\n" );
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document.write( "-kt </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #10522 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=10522\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> YOU ALMOST GOT THE ANSWER..JUST A LITTLE FINISHING TOUCH IS NEEDED..SEE BELOW<BR>\n" );
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document.write( "sin5x+sin3x=4sin2xcos2xcosx<BR>\n" );
document.write( "first i did the regular formula sinA+sinB=2sin(a+b/2)cos(a-b/2)...VERY GOOD<BR>\n" );
document.write( " so it came out like this : 2sin4xcosx......VERY GOOD <BR>\n" );
document.write( "and now i'm stuck and i dont know where to go from that ........WHAT IS THE PROBLEM NOW SEE THE ANSWER AND DO ..WE HAVE COS X WHICH IS OK ..BUT WE HAVE SIN 4X AGAINST SIN 2X COS 2X.....HOPE YOU KNOW THE FORMULA SIN 2X=2 SIN X*COS X<BR>\n" );
document.write( "IT IS DERIVED FROM SIN(A+B)=SIN A*COS B+COS A*SIN B..BY PUTTING A=B..<BR>\n" );
document.write( "SO WE GET SIN 4X=SIN 2*(2X)=2*SIN 2X*COS 2X....HENCE<BR>\n" );
document.write( "SIN 5X+SIN 3X= 2*(SIN 2X) *(COS 2X) *(2*COS X)=4 SIN 2X COS 2X COS X </SPAN>\n" );
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