First off, this particular proof is search bar approved, and does not contain division by zero.
cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2
So where is the error?
EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.
EDIT2: Alright, I asked my calc professor, and he went about saying that I didn't restrict the domain, thus allowed square root problem. Thanks for the help. I also should mention that when evaluating cos pi or sin pi, you need to use radians, not degrees.
And to keep the topic going, what are some other fun math equations you can find?
cos[sup]2[/sup]x=1-sin[sup]2[/sup]x.........................Given
cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup]................Square root each side.
1+ cos x = (1-sin[sup]2[/sup]x)[sup]1/2[/sup] + 1....Add one to each side
1 - 1 = (1-0)[sup]1/2[/sup] + 1..................Evaluate at x = pi (3.14159...)
0 = 1 + 1
0 = 2
So where is the error?
EDIT: I probably should expand, I am trying to find the error, I am not posing this as trivia.
EDIT2: Alright, I asked my calc professor, and he went about saying that I didn't restrict the domain, thus allowed square root problem. Thanks for the help. I also should mention that when evaluating cos pi or sin pi, you need to use radians, not degrees.
And to keep the topic going, what are some other fun math equations you can find?