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!

- anonymous

- katieb

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- anonymous

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?

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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 =)

- anonymous

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

- anonymous

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

- anonymous

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 =)

- anonymous

lol so whos up to what here ?

- anonymous

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

- anonymous

yeh , its relatively easy
lol

- anonymous

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

- anonymous

its a perfect square

- anonymous

you made a mistake between the 2nd last and the last line
a +1 becoame a -1

- anonymous

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

- anonymous

you are doing it the hard way anyway

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ohhh gotcha.

- anonymous

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 )

- anonymous

*facepalm*

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

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