Question

PLEASE HELP!!!
If csc theta= -5/3 and theta has its terminal side in quadrant III, find the exact value of tan2 theta

Answer 1

draw a triangle
it is the mother of all right triangles, the 3 - 4-5 right triangle
from that you can get sine, cosine everything

Answer 2

like this?

Answer 3

I got tan(theta) which is -3/4

Answer 4

I'm not sure how tan^2 actually works...

Answer 5

ok you got \(\tan(\theta)\) what are you looking for ?

Answer 6

for tan^2(theta)

Answer 7

are you looking for \[\tan(2\theta)\] or
\[\tan^2(\theta)\]

Answer 8

i am 100% sure you can square \(-\frac{3}{4}\)

Answer 9

oh wait. it's the first one.

Answer 10

tan2theta. my bad.

Answer 11

\[\tan 2 \theta\]

Answer 12

i never remember that one
is it
\[\frac{\tan(x)}{1-\tan^2(x)}\]

Answer 13

oh no \[\frac{2\tan(x)}{1-\tan^2(x)}\]

Answer 14

I'm not quite sure..

Answer 15

yeah it is the second one
in any case you have the number you need, it is arithmetic from here on in

Answer 16

So I fill x with -3/4?

Answer 17

yes

Answer 18

I got 2tan(-3/4)/1-tan^2(-3/4) = 2.7

Answer 19

ooh NNOOO

Answer 20

lol what?

Answer 21

it is not the tangent OF \(-\frac{3}{4}\) the tangent IS \(-\frac{3}{4}\)

Answer 22

yeah. but I have to find tangent 4 theta not tangent theta

Answer 23

slow

Answer 24

I know tangent theta = -3/4

Answer 25

haha who , me, you or both?

Answer 26

\[\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}\]

Answer 27

in this case
\[\tan(\theta)=-\frac{3}{4}\]

Answer 28

yeah tan2 theta. I keep saying 4 >,<

Answer 29

therefore
\[\tan(2\theta)=\frac{2\left(-\frac{3}{4}\right)}{1-\left(-\frac{3}{4}\right)^2}\]

Answer 30

you don't take the tangent OF \(-\frac{3}{4}\)
the tangent IS \(-\frac{3}{4}\)

Answer 31

i see i made a mistake
you said replace x by -3/4 and i said yes
i should have said HELL NO
you replace \(\tan(x)\) by -3/4

Answer 32

ohh

Answer 33

do I simplify the last equation?

Answer 34

lol it is not an equation it is a number
and yes, no one wants to look at \[\frac{2\left(-\frac{3}{4}\right)}{1-\left(-\frac{3}{4}\right)^2}\] it is just some fraction

Answer 35

I got -24/7

Answer 36

good it is right
there is another famous right triangle with one leg 7 and the other 24

Answer 37

You are very helpiful. Thanks so much :) If you don't mind, would you help me with just one more?

Answer 38

sure hope i don't mess up this time

Answer 39

if 90 degrees < theta < 180 degrees and cos theta=-4/5, find sin 4 theta

Answer 40

crap

Answer 41

LOL

Answer 42

is it complicated?

Answer 43

\[\sin(\theta)=\frac{3}{5}\]

Answer 44

\[\sin(2\theta)=2\cos(\theta)\sin(\theta)\] plug in the numbers let me know what you get

Answer 45

Do I plug in the degrees?

Answer 46

oh no it is like last time
plug in the numbers you know

Answer 47

\[2\times \frac{-3}{5}\times \frac{4}{5}\]

Answer 48

-24/25?

Answer 49

ok so that is \[\sin(2\theta)\]

Answer 50

ooh now I need to get sin4theta

Answer 51

you want \[\sin(4\theta)\] so we repeat the process, but this time with
\[\sin(2\theta)=\frac{-24}{25}\]

Answer 52

sin4theta = 4 * -24/25 * -4/5 ?

Answer 53

\[\sin(4\theta)=2\cos(2\theta)\sin(2\theta)\]

Answer 54

or do I eliminate the negatives?

Answer 55

we have \[\sin(2\theta)\] we need \[\cos(2\theta)\]

Answer 56

ooOh

Answer 57

it is not -4/5
we need the cosine of two theta

Answer 58

would it be? 2sintheta * cos theta

Answer 59

\[\cos(2x)=2\cos^2(x)-1\] so we have to find that as well

Answer 60

do I plug in -24/25?

Answer 61

\[\cos(2x)=2(\frac{16}{25})-1\]

Answer 62

\[\cos(2x)=\frac{7}{25}\]

Answer 63

now plug that in

Answer 64

4 * -24/25 *7/25 ?

Answer 65

no not 4, 2

Answer 66

same formula, different sine and cosine

Answer 67

just change your 4 to a 2

Answer 68

i got -336/625

Answer 69

is that correct?

Answer 70

i calculated 2 * -24/25 *7/25

Answer 71

then it is right

Answer 72

THANK YOU SO MUCH. Sorry for taking so much of your time!

Answer 73

no problem yw