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Let S be the set of all points (x, y) in the coordinate plane such that \(0\leq x\leq \frac\pi 2\) and \(0\leq y\leq \frac\pi 2\). Find the area of the subset of S for which \(\sin^2 x - \sin x\sin y + \sin^2 y\le \frac 34.\)
 

 Aug 2, 2022
 #1
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This is an unusual problem  !!!

 

See the graph here :

 

 

The  feasible area  =  (pi/2)^2  = pi^2/4

 

When x  = 0

sin (0)   - sin (y) + sin^2 (y)  =  3/4

1  - sin(y) +sin^2(y)  = 3/4

sIn^2 (y) - sin (y)  = -1/4

sin^2(y) - sin (y) + 1/4 = 0

(sin y - 1/2)^2  = 0

sin y  =  1/2

arcsin (1/2)  =  y

 

When  y =pi/2   we have (similarly)

sin^2 (x) -sin (x) + 1  =  3/4

sin^2 (x) -sin x = -1/4

sin^2 (x) - sin (x) + 1/4)  = 0

(sin x -1/2)^2 = 0

sin x  = 1/2

arcsin (1/2) = x

 

This sets  up  the area of  the two  right triangles  in the upper left and lower right of the feasible region

The area of these two right triangles  = ( pi/2 - arcsin (1/2)) ( arcsin (1/2))  

 

And the larger right triangle in the upper right corner has the area  (1/2) (pi/2- arcsin (1/2)^2

 

So  the area under consideration  =

 

2.4674  - .5483  - .5483 

 

pi^2 / 4   -  (pi/2 - arcsin (1/2))(arcsin (1/2))  - (1/2)(pi/2 -arcsin (1/2))^2  =   5pi^2  / 36

 

 

cool cool cool

 Aug 2, 2022

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