Numerical trap
consider that there exist three non zero numbers a,b,c
if a,b and c which follow the rule that:
arctan(a/b) = arcsin(a/c) = arccos(b/c)
does it logically follow that a+b > c ?
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3 solutions
Let each of the given expressions be α . Then c o s α = c b = b c . This implies c 2 =1 . Also ( s i n α ) 2 + ( c o s α ) 2 =1. This implies a 2 + b 2 c 4 =1. Or a 2 + b 2 =1= c 2 . Therefore a + b > c
You need to put a backslash "\" before sin and cos as \sin \alpha sin α and \cos \alpha cos α . Note that the function sin and cos is not in italic which is for variables and constants x , y , z , a , b , c . Also note that there is a space between sin and α and cos and α . Only use one pair of \ ( and \ ) for the whole formula. for example: sin 2 α α + cos 2 α = 1 and a 2 + b 2 c 4 = 1 . Notice that the + and = are bigger is LaTex.
What about the case ( a , b , c ) = ( 0 , 1 , 1 ) ?
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Clever observation :)
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Well, I'm shooting myself in the foot - I initially thought the answer was "yes", for the reasons you put - now the answer might be changed to "no", and I'll be marked wrong :-).
if arctan(a/b) = arcsin(a/c) = arcos(b/c) = theta then sin(theta) = a/c , cos(theta) = b/c, tan(theta) = a/b. so this shows that a,b,c are the sides of a right angle trianlge
You have a typo somewhere - in the problem you have b c and in your solution c b .
a r c c o s ( c b ) = a r c c o s ( b c ) ;
take cosine of both sides. you are allowed to do this c = 0 ;
c o s ( a r c c o s ( c b ) ) = c o s ( a r c c o s ( b c ) ) ;
c b = b c
c 2 = 1
c = 1 or c = − 1
as c = − 1 does not make sense → c = 1 ;
This shows that a , b , c are the sides of the triangle. In any triangle, any two sides are always bigger than the other one → a + b > c