Question

Given: costheta= -4/5, sin x =-12/13 ,theta is in the third quadrant, and x is in the fourth quadrant; evaluate cos 2x.


-119/169
119/169
-11/13

Answer 1

you don't need the sine part

Answer 2

then how do I start

Answer 3

Two variables?

Answer 4

\[\cos(2x)=2\cos^2(x)-1\]

Answer 5

oh i see there is a \(\theta\) and an \(x\)

Answer 6

costheta= -4/5
I need cosx

Answer 7

you are asked for \(\cos(2x)\) where \(\sin(x)=-\frac{12}{13}\)

Answer 8

\[\cos(2x)=1-2\sin^2(x)\]

Answer 9

When I do it I get 169

Answer 10

a denominator of 169

Answer 11

there is no \(\theta \) in your question

Answer 12

ok, I'll do it for you, give me a second, this is an easy question

Answer 13

yea I had to put theta for that

Answer 14

satellite, you don't know what you're doing..

Answer 15

\[\cos(x)=1-2\sin^2(x)\]
\[\cos(x)=1-(\frac{-12}{13})^2\]\

Answer 16

there is your 169

Answer 17

\[\cos(2x)=1-(\frac{12}{13})^2\]
\[\cos(2x)=1-\frac{144}{169}\]

Answer 18

\[\cos(2x)=\frac{25}{169}\]

Answer 19

I know, thats what I get, but it isnt a choice

Answer 20

your question asks for \(\cos(2x)\) and doesn't mention any \(\theta\)

Answer 21

I know but that is what is already given

Answer 22

oh damn i made a mistake!! sorry

Answer 23

I'm too lazy to do it
but you need to draw the triangles and either use theta or x for the angle

Answer 24

lol its okay, Ive made plenty

Answer 25

\[\cos(2x)=1-2\sin^2(x)\]
\[\cos(2x)=1-2\times (\frac{12}{13})^2\]
\[cos(2x)=1-2\times \frac{144}{169}\]

Answer 26

\[\cos(2x)=1-\frac{288}{169}\]\[\cos(2x)=-\frac{119}{169}\]

Answer 27

Lol this is a lot of work, can you use this same formula for tan2Theta

Answer 28

nvm. I found the formula, Thank you!