SOLUTION: <font face="symbol">q</font> = sin<sup>-1</sup>(x)+cos<sup>-1</sup>(x)-tan<sup>-1</sup>(x) for the domain x &#8805; 0, what is the range?

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Question 1075749: q = sin-1(x)+cos-1(x)-tan-1(x) for the domain x ≥ 0,
what is the range?
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

q = sin-1(x)+cos-1(x)-tan-1(x) for the domain x ≥ 0,
what is the range?

Because of the inverse sine and cosine functions, 
the domain of q is 0 ≤ x ≤ 1, 

We find the derivative of q



The first two terms cancel and we have



Since the derivative of q is negative, the function q
decreases everywhere on the open part of its domain,
which is 0 < x < 1.
[q is not differentiable at the endpoints 0 and 1, although 
q is defined at those endpoints]

We evaluate q at the endpoints of the 
domain of q : 

Substitute x = 0

q = sin-1(0) + cos-1(0) - tan-1(0) = 0° + 90° - 0° = 90°

Substitute x = 1

q = sin-1(1) + cos-1(1) - tan-1(1) = 90° + 0° - 45° = 45°

Thus the range for q for the domain
 
0 ≤ x ≤ 1

is

45° ≤ q ≤ 90°

Edwin
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