Find the exact value for sec[arctan (-3/5)]

[wiishesssss]

sec[arctan(-3/5)]

cos x = a/h
sec x = h/a

tan^-1 (-3/5) = approx. -31 degrees and lives in quadrant 4
from there i'm fine, and then i know i use the formula a^2 + b^2 = c^2
the thing is how do i know which number is a,b, or c to plus in?

 

[soroban]

Hello,wiishesssss!

There are two possible answers to this problem.

Find the exact value of: \(\displaystyle \,\sec\left[\arctan\left(-\frac{3}{5}\right)\right]\)

Recall that \(\displaystyle \arctan\left(-\frac{3}{5}\right)\) is some angle \(\displaystyle \theta.\)
. . So we have: \(\displaystyle \:\tan\theta \,=\,-\frac{3}{5}\)

Tangent is negative in Quadrant 2 or 4.

Since \(\displaystyle \tan\theta\:=\:-\frac{3}{5}\:=\:\frac{opp}{adj}\)

. . we have: \(\displaystyle \:\frac{opp}{adj} \:=\:\frac{-3}{5}\,\) or \(\displaystyle \,\frac{3}{-5}\)

In both cases: \(\displaystyle \:hyp\:=\:\sqrt{(-3)^2\,+\,5^2}\:=\:\sqrt{3^2\,+\,(-5)^2} \:=\:\sqrt{34}\)


Since \(\displaystyle \sec\theta \,=\,\frac{hyp}{adj}\), we have: \(\displaystyle \,\sec\theta \:=\:\frac{\sqrt{34}}{5}\,\) or \(\displaystyle \,-\frac{\sqrt{34}}{5}\)

 

[wiishesssss]

ah thank you you've been a big help!
do you think by any chance you can help me out with another problem?

 

[tkhunny]

There are two possible answers to this problem.
Tangent is negative in Quadrant 2 or 4.

Let's think about this just for a moment.

Although the second statement is appropriate for the tangent function, that is not the original problem statement. The arctangent function is not usually defined in Quadrant II.

 

[wiishesssss]

The answer is positive because when you do tan^-1(-3/5) = approx. -31 degrees and -31 degrees is in quadrant 4 and in quadrant 4 both secant and cosine are positive therefore your final answer is positive

 

[tkhunny]

The answer is positive because when you do tan^-1(-3/5) = approx. -31 degrees and -31 degrees is in quadrant 4 and in quadrant 4 both secant and cosine are positive therefore your final answer is positive

Well, that isn't exactly evidence. This phrase is questionable, "when you do tan^-1(-3/5)". I'm guessing you mean that you have entered it on your calculator and that's what it said. This is ONLY because your calculator has been defined in this manner. There is NOTHING sacred about this definition. It would be perfectly fine to define arctangent on some other Domain. If you have some compelling reason to deviate from various conventions, feel free to do so and feel responsible to justify your assesment of the situation. In this case, most folks like to avoid discontinuities in the definition. This consideration suggests we avoid \(\displaystyle \pi/2\).