Trig Functions (sin, cosine, tan, csc, sec, cot)

[hlnfan89]

Hello.
I am in 11th grade Algebra II with Trig Honors.
Here is my question:
Sec 140 is equivalent to all of the following except...
a) -sec 40
b) sec (-140)
c) csc 50
d) -csc 50
I said the answer was (b).
That was wrong.
:roll:
I figured out that the right answer was (c), but I had to use the calculator (does anyone feel that we're too dependent on those things?) to figure it out and we weren't allowed to use one on the test...
Also, in my textbook (regents textbook for an honors class... very helpful... not) the questions like this only dealt with numbers like 30, 45, 60, 90, etc. Part of what confused me was that the numbers were 50, 40, and 140. How do you figure out what -sec 40 equals, for example, without the calc?
Please help (and quickly if possible)
Very confused. :?
Thanx,
Laura

 

[ChaoticLlama]

In degree measure, special trig angles are 30, 45, 60, 90 and the like.

This question however cannot be worked out by using such numbers.
Look at reduction identities (or whatever you happen to call them)

You should know that
secx = 1/cosx
cscx = 1/sinx
(just for reference)

cos(-x) = cos(x) -->(because it is symmetric about the y-axis)
cos(90 - x) = sin(x)
cos(90 + x) = -sin(x)
cos(180 - x) = -cos(x)
cos(180 + x) = -cos(x)

I would investigate using these identies (and their sin counter-parts) to solve this question.

Good luck.

 

[hlnfan89]

hmmmmmmmm

OK....
I tried to use the "reduction identities" (that's not what we call them, but I know what you're talking about) and I still got weird numbers...
180-140 is 40 for example. Still not a pretty number :?
90-50 is 40.
Am I using the reduction identities correctly?
I feel that I'm not.
How do you know when to use the 180-x, when to use the 90-x, etc?
Am I being stupid? Is this easier than I'm making it?
Thank you for your time.

 

[Mrspi]

Hello.
I am in 11th grade Algebra II with Trig Honors.
Here is my question:
Sec 140 is equivalent to all of the following except...
a) -sec 40
b) sec (-140)
c) csc 50
d) -csc 50
I said the answer was (b).
That was wrong.
:roll:
I figured out that the right answer was (c), but I had to use the calculator (does anyone feel that we're too dependent on those things?) to figure it out and we weren't allowed to use one on the test...
Also, in my textbook (regents textbook for an honors class... very helpful... not) the questions like this only dealt with numbers like 30, 45, 60, 90, etc. Part of what confused me was that the numbers were 50, 40, and 140. How do you figure out what -sec 40 equals, for example, without the calc?
Please help (and quickly if possible)
Very confused. :?
Thanx,
Laura

sec A = 1/cos A

So,
sec 140 = 1/cos 140

Now, 140 degrees is in the second quadrant, and the reference angle is (180 - 140), or 40 degrees. You need to know that in the second quadrant, cos is negative, so cos 140 = - cos 40. And, sec 140 = 1/cos 140, or 1/(-cos 40). So, sec 140 = - sec 40


You should also know something about cofunctions. Sin x = cos (90 - x), tan x = cot (90 - x), sec x = csc (90 - x), etc. Apply this to this problem.

Can you use this information (WITHOUT the ubiquitous calculator) to answer your question??

In case you haven't figured it out, I am very much opposed to total dependence on calculators for beginning math students. There is a definite place for calculators, particularly in the exploration of the behaviour of functions, etc., but I'm sorry to say that I think calculators have replaced "thinking" for most students today.