anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • 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
  • anonymous
that should be an equal sign...it doesnt look like it on my computer, but it might just be screwy

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
  • 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
  • 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
  • anonymous
i'll try doing the first one*** i'm not sure if i'll be able to
anonymous
  • 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
  • anonymous
lol so whos up to what here ?
anonymous
  • anonymous
i got the second one see if you can do the first one solve for x
anonymous
  • anonymous
yeh , its relatively easy lol
anonymous
  • 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
  • anonymous
its a perfect square
anonymous
  • anonymous
you made a mistake between the 2nd last and the last line a +1 becoame a -1
anonymous
  • anonymous
no, i negated the entire thing, it is NOT a perfect square
anonymous
  • anonymous
you are doing it the hard way anyway
anonymous
  • 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
  • anonymous
then, please, tell us how to do it...
anonymous
  • anonymous
i divided the entire equation by sin^2(x)
anonymous
  • anonymous
ohhh gotcha.
anonymous
  • 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
  • anonymous
*facepalm*
anonymous
  • anonymous
so 0<=x<=4pi ( ie we need two revolutions
anonymous
  • anonymous
trig id overboard.... michelle, listen to elec, he's right
anonymous
  • anonymous
so 2x = 3pi/4 , 7pi/4 ( from the first revolution )
anonymous
  • anonymous
now, we need another revolution, so add 2pi to each of these
anonymous
  • anonymous
so 2x = 3pi/4 , 7pi/4 , 11pi/ 4 , 15pi/ 4 then divide by 2
anonymous
  • anonymous
x = \[x = (3\pi/8 ) , (7\pi/8) , (11\pi/8), (15\pi/8) \]
anonymous
  • anonymous
wow. thank you so so so much. you guys are both like, geniuses. thanks =)) i gave you both medals, whatever that means.

Looking for something else?

Not the answer you are looking for? Search for more explanations.