• sasogeek
Show that 1 + Sin2A = (CosA + SinA)^2
  • Stacey Warren - Expert
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  • chestercat
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  • anonymous
I'll play around with the left side and try to get it to look like the right side.\[1 + \sin2A = \sin^2A+\cos^2A+2\sin A \cos A\]\[=\cos^2A+2\sin A \cos A +\sin^2A\] If we think of sinA as being a and cosA as being b, then the above expression is equivalent to a^2 + 2ab+ b^2, this is the same as the square of the sum a + b... so.... \[\cos^2A+2 \sin A \cos A+\sin^2A = (\cos A+\sin A)^2\]
  • sasogeek
thanks for the response :) I tried to work that out but i got stuck after expanding the first step. thanks again :)

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