Question

tan^2 u = (1-cos2u)/(1+cos2u)
a) use the identities to prove the power-reducting identity
b) explain why this identity does not imply that tan u = square root of (1-cos2u)/(1+cos2u)

Answer 1

reducing**

Answer 2

sen^2(x) = 1/2 - 1/2 cos(2x)
cos^2(x) = 1/2 + 1/2 cos(2x)
so:
tan^2 x=1/2 - 1/2 cos(2x)/ 1/2 + 1/2 cos(2x)

Answer 3

by sen do u mean sin?

Answer 4

ya

Answer 5

if u know what sin^2(u) and cos^2(u) are, u dviide the two

Answer 6

but u dont have sin^2 and cos^2

Answer 7

im confused

Answer 8

tan^2(u) = sin^2(u)/cos^2(u)

Answer 9

you asking to reduse the power.... so now there is no squares

Answer 10

after u say sin/cos though what do u do

Answer 11

oh never mind i get it. how do u answer the second part of question

Answer 12

tan x= sinx/cosx
tan^2x=sin^2x/cos^2x

Answer 13

thats for the second part??

Answer 14

no

Answer 15

then what is part b?

Answer 16

not sure about that

Answer 17

oh . sheena u still there? u know the answer to b?

Answer 18

maybe @experimentX can help, :)

Answer 19

part b

Answer 20

not quite sure .. i'll try.

Answer 21

thanks u have an idea?

Answer 22

i dont think it has to be that complicated tho dont think too hard

Answer 23

@myko have any idea??

Answer 24

maybe becouse of period difference of cos and tan?

Answer 25

I don't think it should make difference .... i was rather thinking this

what if u=0

Answer 26

ya, that looks more scary....:)

Answer 27

the right hand side does not equal the left hand side at the multiples of pi. I think this could be the case.

Answer 28

pi/2

Answer 29

Oo.. what was i thinking.

Answer 30

at pi/2 and 3pi/2 ?

Answer 31

looks like I'm out of my mind ...

Answer 32

huh

Answer 33

something fishy
http://www.wolframalpha.com/input/?i=graph++sqrt%28%281-cos2u%29%2F%281%2Bcos2u%29%29%2C+tan+u

Answer 34

the left hand side is always positive, while the right hand side in not.

Answer 35

http://www.wolframalpha.com/input/?i=graph++sqrt%28%281-cos2u%29%2F%281%2Bcos2u%29%29

Answer 36

ya, i think that's it

Answer 37

it's because we are taking the positive root of a quadratic equation.
It should be rather defined as
tan x = + (1-cos2u)/(1+cos2u) for 0 < u < pi/2 (in interval that is odd multiple of pi/2)
tan x = - (1-cos2u)/(1+cos2u) for pi/2 < u < pi (in interval that is even multiple of pi/2)

Well, that solves it.