4.5 Trig Identities Part 2

Today, we were given a work period to practice proving trigonometric identites. This gave us a chance to work out the more difficult problems with our classmates. One question was especially difficult, and required several people to solve it! It is as follows:

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!