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How do you verify ##(1+sinx+cosx)^2 = 2(1+sinx)(1+cosx)##?

    You will be squaring a quantity that contains both ##sinx## and ##cosx##, so you will need the identity ##sin^2x+cos^2x = 1##. I would start by multiplying both of these out.

    ##(1+sinx+cosx)^2 = 1 + sinx + cosx + sinx + sin^2x + sinxcosx + cosx + sinxcosx + cos^2x##

    ##= 1 + 2sinx + 2cosx + sin^2x + cos^2x + 2sinxcosx##

    while

    ##2(1+sinx)(1+cosx) = 2(1+cosx+sinx+sinxcosx) = 2+2cosx+2sinx+2sinxcosx##

    Compare: ##1 + sin^2x + cos^2x + cancel(2sinx + 2cosx + 2sinxcosx) = 2+cancel(2cosx+2sinx+2sinxcosx)##

    ##1 + sin^2x + cos^2x = 2##

    ##1 + 1 = 2##

    ##2 = 2##

    They’re equal.

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