Posted by Amy on Saturday, March 19, 2011 at 8:57am.
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)
the original problem was:
(sin x + cos x)^2 + (sin x - cos x)^2 = 2
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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