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Arccos(x) + Arcsin(x) = Pi/2


Date: 02/04/2001 at 08:01:13
From: Susan
Subject: Prove 

Prove that for all x between -1 and 1,

     cos^-1(x) = pi/2 - sin^-1(x)

Please help! Thanks.


Date: 02/04/2001 at 10:49:49
From: Doctor Robert
Subject: Re: Prove 

I hope that you've drawn a figure. Use a unit circle and pick a point 
on the circle in the first quadrant. Draw the hypotenuse from the 
origin to this point and then drop the vertical leg to the horizontal 
axis. Let the angle at the origin be a and the other acute angle be 
b. Now label the vertical leg x, the hypotenuse 1 and the horizontal 
leg sqrt(1-x^2).

From the triangle:  

     sin(a) = x    
     cos(a) = sqrt(1-x^2)
     sin(b) = sqrt(1-x^2)
     cos(b) = x

Use the angle-addition formula:

     sin(a + b) = sin(a)cos(b) + cos(b)sin(a)
                = x^2 + (1-x^2) = 1

If sin(a + b) = 1, then a + b = pi/2. (Actually we already knew this, 
since a and b were the acute angles of a right triangle).

But a = sin^-1(x) and b = cos^-1(x).

So...    

     sin^-1(x) + cos^-1(x) = pi/2
     cos^-1(x) = pi/2 - sin^-1(x)

- Doctor Robert, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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