Question

i need to solve these two problem using identities...sin2x+cos2x=0 and csc^2x(x/2)=2secx . help on either or both would be much appreciated!

Answer 1

michelle
i'm not sure if these identifies even exist....
i plugged in values for them and they don't work out.

sin(2pi) + cos(2pi) = 1
though according to the ID you provided, it should be zero

and the second..i'm not sure what it says

is it csc(x/2)^(2x) = 2 secx?

Answer 2

we're supposed to be solving for x using identities. sorry the second is confusing...its supposed to be\[ \csc ^{2}(x/2)=2secx\]

Answer 3

that should be an equal sign...it doesnt look like it on my computer, but it might just be screwy

Answer 4

okay michelle, i kinda gave up on the first and went straight to the second
i really don't think that these exist because i've done it several ways, and I keep getting the same answer
if i got different answers, i might be doing it wrong. but i get the same answer

i'll do it out here and you can decide for yourself
\[\csc^2 ({\frac{x}{2}}) = \frac{2}{\cos x}\]
\[\frac{1}{\sin ^2 (\frac{x}{2})} = \frac{2}{\cos x}\]
\[2\sin^2(\frac{x}{2}) = \cos x\]
\[2(\frac{1 - \cos x}{2}) = \cos x\]
\[1 - \cos x = \cos x\]
\[1 = 2 \cos x\]
\[\cos x = \frac{1}{2}\]

^-- i keep getting that T-T

Answer 5

thats what we're supposed to get! thanks so much...that helps a lot. we're solving for x. and i can solve for cosx=1/2, because we know that x=pi/3 or 5pi/3. thats what we're supposed to be doing =)

Answer 6

AHHHH MICHELLE!!!
THAT MEANS THESE AREN'T IDENTITIES!!!
RAWRRRRRRRRRRRRRRR

you had me tearing my hair out trying to get the equation to equal.. T-T

sighs

give me a bit more time, i'll do the first one

Answer 7

i'll try doing the first one***
i'm not sure if i'll be able to

Answer 8

omg im so sorry. i meant we're solving for x like, using the different identities. idk...thats just what it says on the top of my sheet..."solve for x usiing identities". im sorry...you really dont have to. thanks for that one tho =)

Answer 9

lol so whos up to what here ?

Answer 10

i got the second one
see if you can do the first one
solve for x

Answer 11

yeh , its relatively easy
lol

Answer 12

what i've got so far, i think is right
\[2 \sin x * \cos x + \cos^2 x - \sin^2 x = 0\]
\[2 \tan + 1 - \tan^2 x = 0\]
\[\tan^2 x - 2 \tan x - 1 = 0\]

now you have to solve for x using the quadratic formula, which i dont know how to take into account the tangent

Answer 13

its a perfect square

Answer 14

you made a mistake between the 2nd last and the last line

a +1 becoame a -1

Answer 15

no, i negated the entire thing, it is NOT a perfect square

Answer 16

you are doing it the hard way anyway

Answer 17

wait im confused. how did you get from the first step to the second step (with the tans)? are there identities that tell you that? i might just not know them

Answer 18

then, please, tell us how to do it...

Answer 19

i divided the entire equation by sin^2(x)

Answer 20

ohhh gotcha.

Answer 21

sin(2x) = -cos(2x)

tan(2x) =-1

related angle of pi/4

tan is negative in the 2nd and 4th quadrants

now we originally had 0<=x<=2pi , multiply through by 2 ( because we have (2x) as the argument of our tan )

Answer 22

*facepalm*

Answer 23

so 0<=x<=4pi ( ie we need two revolutions

Answer 24

trig id overboard....
michelle, listen to elec, he's right

Answer 25

so 2x = 3pi/4 , 7pi/4 ( from the first revolution )

Answer 26

now, we need another revolution, so add 2pi to each of these

Answer 27

so 2x = 3pi/4 , 7pi/4 , 11pi/ 4 , 15pi/ 4

then divide by 2

Answer 28

x = \[x = (3\pi/8 ) , (7\pi/8) , (11\pi/8), (15\pi/8) \]

Answer 29

wow. thank you so so so much. you guys are both like, geniuses. thanks =)) i gave you both medals, whatever that means.