Question

Please help with double angle identity!
If siny=-7/23 and y is in Quadrant III, then find the exact value of tan2y

Answer 1

i know tan 2y= 2tan(y) over 1-tan^2(y) but how can I use this identity especially with a sin in there? i know sin is y over r soy is -7 and r is 23...

Answer 2

in the diagram, find cos(y) = ???

Answer 3

cos(y)= square root of 480?! which is 4 square root 30

Answer 4

wait... how can y be in the second quadrant if sin (y) is negative ???

Answer 5

can you double check the question please?

Answer 6

shoot sorry y is in Quad III not II !!!

Answer 7

that makes more sense... thanks... hang on....

Answer 8

yeaaa sorry about that!

Answer 9

would I use the double identity where its like tan(2a)=... or sin(2a)=...

Answer 10

ok... so you found that \(\large cosy=\frac{4\sqrt{30}}{23} \) and you already know \(\large siny=\frac{-7}{23} \)

use the fact tan2y = sin2y/cos2y ....

Answer 11

wait isnt the identity like tan2y= 2tany over 1-tan^2y?

Answer 12

sure you can use that but tany = siny/cosy , which you have....

Answer 13

and tan^2y = sin^2y/cos^2y

Answer 14

but its not looking for tan^2y its asking for tan2y like double angle

Answer 15

yes... the identity is \(\large tan(2y)=\frac{tany}{1-tan^2y} \)

where \(\large tany=\frac{siny}{cosy} \) and \(\large tan^2y=\frac{sin^2y}{cos^2y} \)

Answer 16

sorry... the identity is: \(\huge tan(2y)=\frac{2tany}{1-tan^2y} \)

Answer 17

so...

\(\huge tan(2y)=\frac{2\frac{siny}{cosy}}{1-\frac{sin^2y}{cos^2y}} \)

Answer 18

AFK... be back

Answer 19

hmmm so

Answer 20

i mean