Math

how does sin^2x-cos^2x =1

could someone explain or write down steps on how the left side becomes 1?

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  1. I am positive that you meant that to type
    sin^2x + cos^2x =1

    label a right angled triangle having sides a, b, and c
    with c as the hypotenuse and angle Ø opposite side a.

    LS = sin^2Ø + cos^2Ø
    = (a/c)^2 + (b/c)^2
    = (a^2/c^2) + (b^2/c^2)
    = (a^2 + b^2)/c^2 , but a^2 + b^2 = c^2 by Pythagoras

    so
    LS = c^2/c^2
    = 1
    = RS

    (the choice of variable does not matter, you can call your angle Ø or x or whatever)

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  2. the original problem was:
    (sin x + cos x)^2 + (sin x - cos x)^2 = 2

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  3. In that case you expanded it incorrectly, should have been

    LS
    =(sin x + cos x)^2 + (sin x - cos x)^2
    = sin^2x + 2sinxcosx + cos^2x + sin^2x - 2sinxcosx + cos^2x

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

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