sin2x + sin2y = 2sin(x+y)cos(x-y)

L.S. :

sin2x + sin2y

= 2sinxcosx + 2sinycosy

R.S. :

2sin(x+y)cos(x-y)

= 2(sinxcosy + cosxsiny)(cosxcosy + sinxsiny)

= 2(sinxcosxcos^2y + sin^2xsinycosy + cos^2xcosysiny + sin^2ysinxcosx)

*It appears that this is where many of the student in our class got stuck.*

= 2(sinxcosxcos^2y + sin^2ysinxcosx + sin^2xsinycosy + cos^2xcosysiny)

= 2[sinxcosx(cos^2y + sin^2y) + sinycosy(sin^2x + cos^2x)]

*Similar terms were collected and 'sinxcosx' and 'sinycosy' were factored out.*

= 2[sinxcosx(1) + sinycosy(1)]

= 2sinxcosx + 2sinycosy

Since L.S. = R.S., sin2x + sin2y = 2sin(x+y)cos(x-y).

Remember: when proving trigonometric identities, you must keep an open mind. Don't be afraid to try new things!

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