I am unable to find the answer of this question. Any help will be appreciated

If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.

Are you aware that the graph of $x^2+ y^2= 1$ is a circle with center at (0, 0) and radius 1? So that the circle extends from x= -1 to x= 1 horizontally, and y= -1 to 1 vertically.

Originally Posted by varunkanpur
I am unable to find the answer of this question. Any help will be appreciated

If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.
Technical point: a and A are not the same variable.

-Dan

Assuming you haven't done geometry of the circle you can think of it this way.

$A^2+B^2=1$

$A^2=1-B^2$

$A=\sqrt{1-B^2}$

Think of the maximum and minimum vales of B so that A is real. Then do the same thing for A instead of B.

hmm if you've done trigonometry,try this. Let A=sin x and B=cos x[Parametric equation of circle]. Now range of A+B is Range of sin x + cos x i.e [-1/sqrt(2),1/sqrt(2)].

$sin(x)+cos(x)=\sqrt2sin(x+\pi/4)$
So the range of sin(x)+cos(x) is $[-\sqrt2,\sqrt2]$