Question

find cos(x) given than sin2x = -3/5, 3pi/2>2x>2pi

Answer 1

no clue how to do this=p all help really appreciated

Answer 2

so far im attempting to reason it out like this:

sin2x=2sinxcosx soooo -3/5=2sinxcosx

(-3/5)/2 = -3/10=sinxcosc

Answer 3

You find what 2x is then divide your answers by 2 to find x.

Answer 4

Find cos(2x) That'll get you on the right track :)

Answer 5

\[\large \sin 2x=-\frac{3}{5}\]
\[\large 2x=\sin^{-1} \frac{3}{5}\]
The answers will be in the third and fourth quadrant.

Answer 6

\[\huge \cos(2x) = \sqrt{1-\sin^2(2x)}\]

Answer 7

And then divide by 2 to find x-calues.

Answer 8

although I'm betting cos(2x) = 4/5
or -4/5

Answer 9

but then again, you specified a quadrant, so cos(2x) is 4/5 after all? :D

Answer 10

oh wait...
it's -4/5

Answer 11

man you guys are fast! thanksfor all the help so far!! working on finding cos2x

Answer 12

Okay, I'm dizzy...

Answer 13

haha! so i found that cos2x is 4/5, but are you allowed to divide cos2x by 2 in order to find cosx?

Answer 14

\[\huge \frac{3\pi}{2}<2x<2\pi\]
it must be this interval... then cos(2x) must be positive.

Answer 15

or maybe its a formula i need to use=p gotta check

Answer 16

\[\large 2x=216`52', 323`8'\]
Using that restriction.
\[\large 2x= 323`8'\]
This ` symbol denotes a degree sign. Don't have it on my keyboard.
Then
\[\cos 2x=\cos 2(323`8')\]

Answer 17

silly @mrdrew445
\[\huge \cos(x) =\sqrt{\frac{1+\cos(2x)}{2}}\]
^.^

Answer 18

Half angle identities are the way to go ^.^

Answer 19

ooooooo i didnt even realize! i didn't think i could put a double angle in there because it just says "1+cos(theta)"

duhhhh lol

Answer 20

Oops made an error
\[\large \cos2x=\cos 323`8'\]
And @PeterPan you're going along the long road. This question always gets students to solve it by using half angle identities etc. This is just using logic.

Answer 21

and a calculator, @Azteck
which is likely not at hand ^.^

Answer 22

The answer is
\[\frac{4}{5}\]

Answer 23

That's only if it was cos2x. Your question says cosx

Answer 24

^nope
\[\huge \frac{3\pi}{2}<2x<2\pi\]\[\huge \frac{3\pi}{4}<x<\pi\]
and cosine is negative in this interval ^.^

Answer 25

You could get cos(x) without a calculator. I'm quite sure the teacher was shooting for the students to use identities on this one
like I said\[\huge \cos(x) =\sqrt{\frac{1+\cos(2x)}{2}}\]
and you already know what cos(2x) is

Answer 26

sooo i ended up getting 3/5, but that doesn't seem right, gunna do it over to check

Answer 27

thank you very much peter and azteck!!!

Answer 28

just to make sure... cos(x) should be negative, ok?
\[\huge \cos(x) =-\sqrt{\frac{1+\cos(2x)}{2}}\]

Answer 29

because x ends up in an interval where cosine is negative

Answer 30

Has he even learnt half angle identities. QUestions like this usually come apparaent once you learnt he first few identities.

Answer 31

apparent*

Answer 32

the*

Answer 33

i just recently learned half-angle identities, a few days ago

Answer 34

There you have it. This is a logical test of knowledge... of half-angle identities
^.^

Answer 35

Okay then Peter Pan's method is the way to go.

Answer 36

To be honest, I've never learnt half angle results through my years. So my way was because I lacked knowledge of half angle results. Our syllabus does not require you to know half angle results.

Answer 37

Half-angle identities are a pain ^.^
and not as useful as the others, it would seem :D

Answer 38

all these identities and proving identities i need alot more work on! i find them all confusing=\ and im set to be taking calculus next year i must master them haha!

Answer 39

@PeterPan That's basically why it isn't in our syllabus. LOL

Answer 40

Yes~ you must^.^

Answer 41

thanks peter appreciate it alot. fairly certain i have the right answer now (-3sqrt10/10)