Question

Evaluate the following. Answers must be exact.

a. -6/7
b. -3/ 2sqrt2 or -3 sqrt2/4
c. 1/ 5sqrt13 or sqrt 13/65

Answer 1

calculator?

Answer 2

no, it doesnt work if you plug it all in, you need exact values

Answer 3

have you tried it in both degree mode and radian mode?

Answer 4

yes

Answer 5

they need to contain sqrts and stuff.

Answer 6

2/9 for first one.

Answer 7

http://www.wolframalpha.com/input/?i=tan%28arctan%284%29-arcsec%28%28sqrt%285%29%29%29&dataset=

Answer 8

be in radians

Answer 9

no the answer to the first 1 is -6/7, i have the answers but i dont know how to solve them

Answer 10

that bottom part is it over that interval. or is that the start of new question?

Answer 11

thats part of a different question

Answer 12

Well for first one
tan(arctan(4)-arcsec((sqrt(5)))

i think the tan and arc tan will cancel. I say express it all in terms of tan. Maybe lol.

Answer 13

I have no idea, that's why I asked for help

Answer 14

Yes those are tricky ones. Trig is not my strength. You must really know your trig IDs :)

maybe I am wrong but isn't there a

tan(a-B)
and it
= tan[a (-B) ]

Answer 15

uhhhh no, tan (x-y)= (tan x-tan y) / 1 + tan x tan y)

Answer 16

I solved b.....

Answer 17

Great job :D lol

Answer 18

i dont know how i did that..... but now i cant solve a or c

Answer 19

It is hard. BUt you need to find the right trig ID :P

Answer 20

gah

Answer 21

I'm not buying that the answer to the first problem is -6/7. I believe it is 2/9.

Here are the two triangles: (left one is for ArcTan, right one for ArcSec)

Answer 22

thats the key that i was given, by my professor :/

Answer 23

Maybe both same on a different interval. Trig functions are repeating btw.

Answer 24

I know they are repeating, but it doesn't make it any easier....

Answer 25

well if I do it that way, how do i convert .2222222 to a fraction?

Answer 26

Here's how I get it symbolically: use \[\tan(u\pm v) = \frac{\tan u \pm \tan v}{1\mp \tan u \tan v}\]\[u = \tan^{-1}(4),v=\sec^{-1}(-\sqrt{5})\]

Answer 27

will you work me through each problem then... ?

Answer 28

im not sure where you got 4 and 2

Answer 29

Look at the triangles I gave you. We know that the tangent of the angle in the first one = 4, so the opposite side /adjacent side must be 4. We arbitrarily set the adjacent side to be 1, and the opposite side to be 4, then use the Pythagorean theorem to find the hypotenuse. The sides just have to be in the right ratio, which they are. Now it is trivial to find \(\tan^{-1}(\tan(\theta))\) anyhow because you just find the right angle in the 1st quadrant, right? But doing the triangle allows us to compute other trig functions, like \(\tan^{-1}(\csc^{-1}(\theta_2))\) where \(\theta_2\) is the angle in the second triangle.

Answer 30

there is no way, i could make my own triangles by myself....

Answer 31

i don't even understand where you got them

Answer 32

Sure you could. You just have to remember SOH CAH TOA

sin is opposite over hypotenuse
cos is adjacent over hypotenuse
tan is opposite over adjacent

(and I suppose you need to remember that sec = 1/cos, csc = 1/sin, cot = 1/tan)

Answer 33

yeah i know those identities

Answer 34

right. well, that's all I did! we know that the tangent of some angle = 4, so that means that opposite / adjacent = 4 which means opposite = 4* adjacent. I just drew a right triangle where that was true, and then solved for the hypotenuse (which I didn't actually even need here, but no matter).

Answer 35

that just confuses me....

Answer 36

for the other triangle, we knew that sec of the angle = -sqrt(5) but sec (-u) = sec (u) because cos(-u) = cos(u), right?
so sec of the angle = sqrt(5), and sec is just 1/cos, so 1/cos of the angle = sqrt(5), and cos is adjacent over hypotenuse so 1/cos = hypotenuse/adjacent. we set adjacent to be 1, and hypotenuse to be sqrt(5). Pythagoras rides to the rescue again and we find the other side to be (sqrt(5))^2 = 1^2+b^2 -> 5 = 1+b^2 -> 4 = b^2 ->b=2

Now we can compute any trig function on that angle by plugging in the lengths of the sides of that triangle in the basic formulas. tan of that angle is just 2/1 = 2.

Answer 37

Here is what I came up with using graphs....

Answer 38

Here is a slightly more coherent approach

Answer 39

okay, thanks

Answer 40

I forgot to account for the secant being negative, which changes the picture entirely :-( My apologies to the nameless prof who did have the correct answer after all. When I saw @timo86m's Wolfram link I didn't notice that he'd mistyped the expression and it was corrected to something else, so when I made a sign error in doing it numerically and got the same answer, I assumed it was correct. My bad!

Answer 41

Properly executed, it would have been
\[\tan u = 4\]\[\tan v = -2\]
\[\tan(u-v) = \frac{\tan u - \tan v}{1+\tan u \tan v} = \frac{4-(-2)}{1+(4)(-2))} = -\frac{6}{7}\]