Question

Which of the following is not an identity?

A. cos^2 x csc x - csc x = -sin x

B. sin x(cot x + tan x) = sec x

C. cos^2 x - sin^2 x = 1- 2sin^2 x

D. csc^2 x + sec^2 x = 1

***i don't really get this one :( @terenzreignz? what do u think?

Answer 1

ughh.. last one, and i don't get it :(

*yells at brain* :P

Answer 2

Which one does your instinct dictate? :D

Answer 3

lol, well earlier, my instincts took me to the wrong place haha, so let's give it a try again this time... :p

I think it's either A or C (again hahah looks like my instincts have quite the distinct pattern lol)

but I'm leaning towards A... can you have a -sin x at the end of the equal sign? :/

Answer 4

I see... I'll let @sriramkumar have his say... ;)

Answer 5

this is the second chance i was talking about hahah

question is, can the feelings/instinct redeem themselves?

Answer 6

kk :)

Answer 7

yep its the last once since we have csc^2x = 1/sin^2x and sec^2x = 1/cos^2x


then you'll have it

1/cos^2x + 1/sin^2x = 1

cos^2x + sin^2x / cos^2x sin^2x = 1

1=cos^2x.sin^2x

and its true for only some or none values of x, so it can't be an identity since an identity should be true for all values of x.

hope u understood

Answer 8

Proving something isn't an identity... well, if you could find a value of x which would make an equation false, then that equation is not an identity...


Try having \(\huge x = \frac{\pi}{4}\) for letter D

Answer 9

or simply say x=0... put it in all equations and see which one doesn't satisfy??

Answer 10

That won't do it, @sriramkumar :P

Answer 11

Or on second, thought, it will :D My bad...

Answer 12

so you get 1=1 right?

Answer 13

??? @terenzreignz an equation has a limited set of satisfying values, and an identity is true for all values of the x :P so try substituting 2 or 3 values of x then that would do

Answer 14

Hardly... when
\[\huge \sec\left(\frac{\pi}{4}\right)=\csc\left(\frac{\pi}{4}\right)=\sqrt2\]

Answer 15

okay, so then you'd get 2=1 ? :/ cuz sq.rt. 2 + sq.rt. 2= sq.rt. 4 which equals 2 right?

but if thats the case, then this isn't possible right? cuz 2 does not equal 1... :/ but maybe i did something wrong... which is quite likely :/

Answer 16

so you want me to solve and show all the cards in there??? I can say that all the first three can be solved on for RHS but not in 4, you can't get RHS

Answer 17

what's RHS? :/

Answer 18

just Right Hand SIde :P :D :/

Answer 19

haha ohh oaky i see :P

Answer 20

still you'vent understood?

Answer 21

In any case, @iheartfood , letting
\[\large x = \frac{\pi}{4}\] should do it for choice D

However...
\[\Large \sqrt2 + \sqrt2 + \sqrt 4\]
????

Maybe you need more ice cream :P

Answer 22

Besides... the csc and the sec were SQUARED in equation D.

Answer 23

so since that's the case, D is the one that isn't an identity? cuz you're supposed to get 1=1 right? but here, we got 4=1? @terenzreignz does the thingy you put above get 4?

Answer 24

Yes, it does get 4, @iheartfood
And yes, D is the one which isn't an identity :P

Answer 25

lol :p

this should be enough ice cream :)

http://www.google.com/search?newwindow=1&hl=en&q=kitchen+sink+ice+cream&bav=on.2,or.r_qf.&bvm=bv.45960087,d.cGE&biw=1280&bih=680&um=1&ie=UTF-8&tbm=isch&source=og&sa=N&tab=wi&ei=BgmCUdVls4yKAsqkgMAD

if only it were real... *sigh*

Answer 26

yay!! :) so in order for it to be an identity, it'd have to be 1=1 right?

Answer 27

No, for it to be an identity you have to prove it, by turning the LHS into the RHS
using some manipulations...


And you can't do that in D.

Answer 28

ahhh i see.. :)

so as long as they equal each other in the end?

Answer 29

Yes.
Proving identities takes some creativity. Would you like me to demonstrate?

Answer 30

yes please! :) an example would be great to help clear things up! :)

Answer 31

Which one? A B and C are identities, so which one do you want me to prove...?

Answer 32

can you please do A? :) and possibly C after if you have time? :)

Answer 33

Well okay.
First, let's start with the Pythagorean identity, you know this, right?
\[\huge \sin^2(x) + \cos^2(x) = 1\]

Answer 34

mhm :)

Answer 35

Then it follows that \[\huge \sin^2(x)= 1-\cos^2(x)\]


and
\[\huge \cos^2(x) = 1-\sin^2(x)\]



right?

Answer 36

I'm following :)

Answer 37

So from the first one, it also follows that
\[\large -\sin^2(x)=-[1-\cos^2(x)]=\cos^2(x)=1\]

catch me so far?

Answer 38

yuppp :)

Answer 39

rats... I meant
\[\large -\sin^2(x)=-[1-\cos^2(x)]=\cos^2(x)-1\]


\[\huge -\sin^2(x)=\cos^2(x)-1\]

Answer 40

Okay?

Answer 41

lol.... okay :)

Answer 42

Now, to proceed with A.
We start with the LHS (Left-Hand side)
\[\large \cos^2(x)\csc(x) - \csc(x)\]

Answer 43

Now, since
\[\large \csc(x) = \frac1{\sin(x)}\]

The LHS may be written as...
\[\large \frac{\cos^2(x)}{\sin(x)}-\frac1{\sin(x)}\]
Catch me so far?

Answer 44

yeah making sense :)

Answer 45

Now, since they have the same denominator, it can be written as
\[\large \frac{\cos^2(x)-1}{\sin(x)}\]

Answer 46

ahhh i see :)

Answer 47

Now, the numerator can be simplified... into what? (refer to the preliminary identities I showed you :) )

Answer 48

okay, hmmm tricky tricky! :P

urmm do i still need to include the cos ? ;/ or does it change to something else?

Answer 49

Scroll up, we derived an equation... using the Pythagorean identities... remember?

Answer 50

lol its like you explain, and i can follow cuz its there! but then i have to fill in the blank parts, and its confusion and chaos! at least in my brain haha gotta work on this stuff ;/

Answer 51

ohhh i see,,,

so it simplifies to this? sin^2 x ?

Answer 52

Are you sure? Check it again...

Answer 53

urmm.... lemme see.. i got that from this part sin2(x)=1−cos2(x) ^^^

but does it not apply like that? where i can just flip the order around? :/

Answer 54

I took a screen-cap... just to make a point :P

Answer 55

ohh i see.. so it simplifies to -sin^2(x) ?

Answer 56

Yes. For heaven's sake, these things I tell you, don't forget them :P

\[\huge \frac{-\sin^2(x)}{\sin(x)}\]

Which obviously simplifies into...?

Answer 57

-sin(x) ?

Answer 58

Yes. And as you can see, we have (magically) turned the LHS into the RHS (Right-hand side)


Et voila
I guess :P

Answer 59

lol wow! i see :)

thanks!!! :D

Answer 60

Now, C.

Answer 61

kk :)

Answer 62

Start with the LHS as usual...
\[\large \cos^2(x) - \sin^2(x) \]

Now, we can express \(\cos^2(x)\) in terms of \(\sin^2(x)\)

Once again, into what?

Scroll up again, see the alternate way we can express \(\cos^2(x)\)

Answer 63

this part right?

−sin2(x)=−[1−cos2(x)]=cos2(x)=1

Answer 64

No.

Answer 65

You said you could follow :/

Answer 66

yeah =(

still a bit confused tho :/

it is also this right? 1−sin^2(x)

Answer 67

Yes... again, to make a point :P
(you DID say you were following...)

Answer 68

Anyway...
\[\Large \cos^2(x) - \sin^2(x) \]You can replace the \(\cos^2(x)\) with...?

Answer 69

haha yeah i recall that :)

Answer 70

just got a bit lost just now :) recovered!


so i replace it with 1-sin^2 x ?

Answer 71

Okay.
\[\Large 1-\sin^2(x) - \sin^2(x)\]
Simplifying yields...?

Answer 72

1-2 sin^2 x ?

Answer 73

Which is the RHS.

Once again, we have managed to turn the LHS into the RHS... as if by magic :P

This was basically an overview.. better consult your instructor on proving identities :)

That was fun, no? :P

Answer 74

haha yeah! its like a miracle! :P lol and yeah i will! thank you sooo much!! :)

and lol super fun!! minus the parts where i experienced short-term memory and temporarily got lost hahaha :P

yayy!! :) thank you so much for helping me @terenzreignz :)