sasogeek
  • sasogeek
Show that 1 + Sin2A = (CosA + SinA)^2
Mathematics
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sasogeek
  • sasogeek
Show that 1 + Sin2A = (CosA + SinA)^2
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • 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
  • 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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