There's a trigonometric identity in (sin A)^2 + (cos A)^2 = 1. But just saying it doesn't prove anything so allow me to use basic geometry to explain to those who have no clue what I'm saying. You can see this by knowing "sohcahtoa". It's a mnemonic device of sorts used to remember the equations used to calculate the following trigonometric functions:
sin A = opposing side / hypoteneuse (or s=o/h)
cos A = adjacent side / hypoteneuse (or c=a/h)
tan A = opposing side / adjacent side (or t=o/a)
[Note: And hence, s=o/h, c=a/h, and t=o/a yield "sohcahtoa".]
And of course, Pythagoras' Theorem (h^2 = o^2 + a^2). So since A=B and C=90, sin 90 = 1 and from...
sin(A)sin(B)sin(C) + cos(A)cos(B) = 1
...we're left with:
sin(A)sin(A)*1 + cos(A)cos(A) = 1
The multiplied 1 is omiited, and similar tems are grouped together:
(sin A)^2 + (cos A)^2 = 1
There ^2 means the preceding term is squared. So substituting our trusty "sohcahtoa" equations:
(o/h)^2 + (a/h)^2 = 1
Then square above and below, then group into one fraction:
(o^2 + a^2)/h^2 = 1
Insert Py's theorem here...
h^2/h^2 = 1
And so, cancelling h^2 above and below yields a beautiful...
1=1
There you go. Proven.
(Note: For something as useless to my profession as trigonometric functions are, I actually like playing around with them...)
Edited on August 21, 2006, 1:05 am
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Posted by Alexis
on 2006-08-21 00:58:11 |
My solution
