Proof of a^2+b^2=1: Step-by-Step Guide

[Wildcat]

given a*(1-b^2)^1/2 +b(1-a^2)^1\2 =1 prove a^2 + b^2 =1

I tried squaring both sides and then squaring again to get
a^4 + b^4 -2b^2 -2a^2 +2a^2b^2 +1 =0

and that could be (a^2 + b^2)(a^2 + b^2) - 2(a^2 + b^2) = -1

I don't know where to go from there and not sure this is even correct.

Can someone help?

 

[mfb]

With c=a^2+b^2, your equation becomes c^2-2c=-1. This has just one solution for c...

I moved your thread to our homework section, as it looks like a homework question (at least it is very similar to them).

 

[WWGD]

Try completing the square, i.e., try to express in the form $$(x+y)^2$$ . If you use

double $'s ( the signs, to do Tex--tho having general $'s
will always help in daily life too -- you can easily Tex your expressions, e.g:

(a+b)^2 , using

Double$'s

(wrapping around) will get you

$$(x+y)^2$$ .

 

[Infrared]

If you don't mind using some trig...

Since -1 \leq b \leq 1 and -1 \leq a \leq 1 we can make the substitutions a=sin\alpha and b=sin\beta. You should get that if a=sin\alpha, then b=sin(\pi/2-\alpha)= cos\alpha which solves the problem. This avoids a lot of algebra.

 

[Wildcat]

With c=a^2+b^2, your equation becomes c^2-2c=-1. This has just one solution for c...

I moved your thread to our homework section, as it looks like a homework question (at least it is very similar to them).

OK so $$c^2 -2c +1 =0$$
then (c-1)(c-1)=0 implies c=1 so
$$a^2 + b^2 =1$$

Right??

 

[WWGD]

ε



$$(a+b)^2$$ TEST

Right, but I meant to say that you need to choose both a,b here to fit your formula.

Sorry first-of-all for my poor choice of letter a,b here. I meant you could rewrite your formula

as a sum $$(x+y)^2 $$, with just the right choice of x,y.

 

[mfb]

OK so $$c^2 -2c +1 =0$$
then (c-1)(c-1)=0 implies c=1 so
$$a^2 + b^2 =1$$

Right??

Right.

Please use the edit button for code tests.