Posted by **Amy** on Saturday, March 19, 2011 at 8:57am.

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

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

- Math -
**Reiny**, Saturday, March 19, 2011 at 9:04am
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)

- Math -
**Amy**, Saturday, March 19, 2011 at 9:06am
the original problem was:

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

- Math -
**Reiny**, Saturday, March 19, 2011 at 12:04pm
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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