## anonymous 5 years ago 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!

1. 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?

2. anonymous

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

3. anonymous

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

4. 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

5. 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 =)

6. 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

7. anonymous

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

8. 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 =)

9. anonymous

lol so whos up to what here ?

10. anonymous

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

11. anonymous

yeh , its relatively easy lol

12. 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

13. anonymous

its a perfect square

14. anonymous

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

15. anonymous

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

16. anonymous

you are doing it the hard way anyway

17. 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

18. anonymous

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

19. anonymous

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

20. anonymous

ohhh gotcha.

21. 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 )

22. anonymous

*facepalm*

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. anonymous

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