On the other hand, when we let y range from pi to 3/2pi or 180 to 270, then we have sin y / sec^2(y) and sec^2(y) is always positive but sin y is negative when y is between 180 and 270 degree's so the slope in this case is negative.
So in the first case, you can find that sin y / sec^2 y is equivilent to x sqrt(x^2 - 1) within the limits of the range, and in the second case sin y/sec^2 y is equivilent to |x| sqrt (x^2 - 1) within the limits of the range.
This is why I feel trigonometric identities are extremely dangerous. I'm always uncomfortable when using them...
]]>Below is a graph of x as a function of y x = sec (x)
to make this into a function we use the first range restriction: [0, pi/2) and
[pi, 3/2pi)
Now using the alternate definition where the range restriction is [0, pi/2) and (pi/2, pi] we have:
Note the derivatives. It makes sense the second derivative contains the absolute value term since we can see from the graph the slope is always postive, but how would we figure that out without graphing?
To calculate the derivaties, we have y = sec^-1(x) so sec (y) = x differentiating implicitly we get dy/dx = 1/(sec y tan y) now here's where I start to get uncomfortable. We know sec(y) = x and must be x. But if we say tan(y) = sqrt( 1 - sec(y)) or tan y = sqrt ( 1 -x^2) we are really finding |tan(y)| therefore when tan (y) < 0 the derivative should be incorrect. So why isn't it?
In the second equation we have to replace sec(y) with |x|. How come? y ranges between 0 to pi/2 or 0 to 90 in which case the secant IS positive, but also from pi/2 to pi or 90 to 180 degree's, where the secant while be negative.
This is rather confusing...
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