Quick Trigonometry Q

[113zami]

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find the value of tan(arccos2/3)?

for the correct ansr, they say "let u=arccos2/3, then cosu=2/3. Because
cosu is (+), u is a 1st quad angle."

but why can't u be a 4th quad angle, isn't cosu also (+) in the 4th quad (that is, if you take the counterclockwise angle)??

for example cos(pi/4)=sqrt2/2 and cos(7pi/4)=sqrt2/2 also

so i don't see why can't u be a 4th quad angle???

thanks

 

[Streetwolf]

find the value of tan(arccos2/3)?

for the correct ansr, they say "let u=arccos2/3, then cosu=2/3. Because
cosu is (+), u is a 1st quad angle."

but why can't u be a 4th quad angle, isn't cosu also (+) in the 4th quad (that is, if you take the counterclockwise angle)??

for example cos(pi/4)=sqrt2/2 and cos(7pi/4)=sqrt2/2 also

so i don't see why can't u be a 4th quad angle???

thanks

In order for arccos to be a function, it has to pass the vertical line test (graph it). When you graph cos(x) you get an oscillating function between y = [-1, 1] that stretches from x = (-inf, inf). If you take arccos(x), you'd have an oscillating 'function' between x = [-1, 1] that stretches from y = (-inf, inf). This can't happen with an actual function. For any given x between [-1, 1] you'd have infinitely many y values. So the powers that be decided that arccos(x) is to be bound between y = [0, pi]. So any given x between [-1, 1], when plugged into arccos(x), has a value between [0, pi]. And since 2/3 is positive, you are in quadrant 1.

 

[113zami]

so does that mean that arcsinx and arctanx can only be in the 1st and 4th quads?

 

[Streetwolf]

I think arctan hits up -pi/2 to pi/2 (1st and 4th) and arcsin hits up -pi/2 to pi/2 as well. Graph them, you'll see where the repeats are. Basically you'll see the same function stacked on top of each other bounded by two values. It's easiest to visualize with arctan since the functions are "separated" from each other, since they head off to infinity and -infinity.