Question

Trig question
http://imgur.com/QJXLSTh
I'm not so sure what the question means so any help would be appreciated!

Answer 1

Looks like you just have to modify those using identities and angle sum formulas etc... then change tanx into k wherever possible

http://www.regentsprep.org/regents/math/algtrig/ATT14/formul30.gif

use that to simplify tan(pi-x)... then replace all tanx's with k.

Answer 2

I'm not so sure how to make use of those formulas

Answer 3

keep in mind that, the angle "x" is ACUTE, and the tangent function is POSITIVE, tangent is positive only where "x" and "y" are the same sign, negative or positive both, the only happens in the 3rd Quadrant and 1st Quadrant
so is less than \(\cfrac{\pi}{2}\)

Answer 4

I'm completely confused on what to do...

Answer 5

you're supposed to find angle "x" in short

it's tangent, which is "k", which is just a value,, doesn't matter what it might be, what matters is that is positive, so the tangent is positive, that only occurs where "x" and "y" are both positive, or both negative, that only happens in the 1st and 3rd Quadrant, check your Unit Circle, if you don't have one, do get one, many good ones online

so we know that tangent is positive, so is in the 1st or 3rd Quadrant, and we also know that is acute, from there, check your choices, which one will give you an angle in the 1st or 3rd Quadrant that's acute?

Answer 6

i posted the wrong formula
http://www.regentsprep.org/regents/math/algtrig/ATT14/formul56.gif
\[\Large \tan (\pi- x) = \frac{ \tan \pi - \tan x }{ 1+\tan \pi \tan x }\]

Remember that tanx=k... ie... replace tanx with k.

Answer 7

@jdoe0001 i don't think we're finding x. We just want to write those things in terms of k.

Answer 8

hmmm... ohh I thought those were choices =)

Answer 9

\[\Large \tan (\pi- x) = \frac{ \tan \pi - \tan x }{ 1+\tan \pi \tan x }\] so @williamnz simplify this, knowing that tanx=k, and i'm sure you can find the value of tan(pi)

Answer 10

ok... so we know that tan(x) = k

so lemme do the 1st one
\(\bf tan(\pi-x)=\cfrac{tan(\pi)-tan(x)}{1+tan(\pi)tan(x)}\\ \quad \\
tan(x) = k\qquad \qquad thus\\ \quad \\
tan(\pi-x)=\cfrac{tan(\pi)-tan(x)}{1+tan(\pi)tan(x)}\implies tan(\pi-x)=\cfrac{tan(\pi)-k}{1+tan(\pi)k}\)

Answer 11

In the answers it says that tan(pi-x) = -k
But I can't simplify that into -k

Answer 12

What's tan(pi) equal to?

Answer 13

0

Answer 14

You can simplify it then.

Answer 15

0-k/1+0=-k I presume?

Answer 16

Yep.

Answer 17

So I just need to use those formulas?

Answer 18

Yeah try using it again for the second one, use the same formula.

Answer 19

\(\bf tan(\pi-x)=\cfrac{tan(\pi)-k}{1+tan(\pi)k}\\ \quad \\
\color{blue}{tan(\pi) = \cfrac{sin(\pi)}{cos(\pi)}\implies \cfrac{0}{-1}\implies 0}\\ \quad \\
tan(\pi-x)=\cfrac{tan(\pi)-k}{1+tan(\pi)k}\implies tan(\pi-x)=\cfrac{0-k}{1+0\cdot k}\implies \cfrac{-k}{1}
\)

Answer 20

Thank you very much! For part ii however, when I do tan(0.5pi) I get math error?

Answer 21

Yeah i was wondering about that too... tan(0.5pi) is undefined.

Answer 22

The answers say part ii = 1/k...

Answer 23

Maybe we can try breaking it up using this way \[\Large \tan(0.5 \pi - x) = \frac{ \sin (0.5 \pi - x) }{ \cos (0.5 \pi - x) }\]

Answer 24

Then use http://www.regentsprep.org/regents/math/algtrig/ATT14/formul30.gif for both numerator and denominator

Answer 25

I think that's way too much working for 1 mark, there must be an easier way.

Answer 26

Not sure that'll help, though :/ since we need tanx=k and that gets rid of tan entirely

Answer 27

Hmmm what about part iii, the answer notes say something about a 90 degree triangle

Answer 28

Yeah maybe we can do that for part 2... we know the angle is acute

Answer 29

boom. Part 2 and 3.

Answer 30

I see... Thanks so much for your help!
I don't see why there's so much working for 4 marks in total when other questions which take like 2 minutes give you like 5 marks or more...

Answer 31

Well you can do 2 and 3 super quick with a triangle. 2 and 3 together took about a minute once i made the triangle.

Answer 32

Hmmm but still we haven't learnt any of this in class lol

Answer 33

none of the angle sum formulas? or the triangle method? Idk if it's possible to do part 1 with a triangle... if the angle is acute, then pi-x is obtuse...and right triangles don't have obtuse angles

Answer 34

Nope though I guess the triangle method is general trigonometry knowledge