Question

Help? Will fan and medal?
Can someone help me with these five problems, or at least the first one to help explain how to do these?
For questions 1-5 label the equation as (a) an identity or (b) not an identity.
1. sin^2 x + cos^2 x = 1. (I think this is a not sure.)
2. sin x + cos x = 1 (is this not one? so b?)
3. sec x + csc x = 1 (I'm not sure)
4. cot^2 x = 1/tan^2 x (unsure..)
5. csc^2 x -1 = cot^2 x (Unsure..)
If someone can help I'd really appreciate it.

Answer 1

1) is definitely true, it is the mother of all trig identities

Answer 2

2) is definitely not an identity
you can check by plugging in some number for x and see that you don't get 1

Answer 3

@satellite73 would you mind explaining how that is proven? can you plug in a variable to get the answers?

Answer 4

i think i can

Answer 5

Okay! Thank you SO MUCH.

Answer 6

if you plug in a number and don't get the answer, then it is surely not an identity
an identity means it is true for all x

Answer 7

proving that it is an identity is a totally different story though

Answer 8

As in?

Answer 9

are you used to working with degree, or radians?

Answer 10

A little

Answer 11

i meant which ones do you use usually? \[\sin(\frac{\pi}{4}\]or \[\sin(45^\circ)\]

Answer 12

She has had us work with both, actually. I am just not a genius in it as I am in my advance algebra. Haha

Answer 13

ok well then lets just say if you plug in a number (don't use 0 or 180) and you don't get the answer, then it is not an identity

Answer 14

now how about \[\cot^2(x)=\frac{1}{\tan^2(x)}\]

Answer 15

do you know the definition of cotangent?

Answer 16

not precisely.. all I know is it is A/O.

Answer 17

ok that will work

Answer 18

and tangent, to use your language, would be O/A

Answer 19

I remember that from SOA CAH TOA AH.

Answer 20

i.e. cotangent is the reciprocal of tangent, and vice versa

Answer 21

That's the only way I have survived trig up to this point. I really don't like trig. And okay, gotcha!

Answer 22

A simple algebraic example of an identity would be 4x + 10x = 14x.

No matter what number you chose for x the LHS = RHS

Answer 23

so \[\cot^2(x)=\frac{1}{\tan^2(x)}\] is just like saying \[\left(\frac{1}{z}\right)^2=\frac{1}{z^2}\] which is ture

Answer 24

*true

Answer 25

I assume "z" is the varriable you used for cot and tan

Answer 26

yes i was using z for tangent

Answer 27

Okay. It is basically showing that since tan is recipcrical to cot, that ratio you shows represents the fucntion between cot and tan, so that statement is then true. if that made sense.

Answer 28

to write it all out, \[\cos^2(x)=\left(\frac{1}{\tan(x)}\right)^2=\frac{1}{\tan^2(x)}\]

Answer 29

yes, what you said

Answer 30

Okay, I got you. Now what if there was no squared variables, would it still apply since cot is always equal to the recip of tan?

Answer 31

i made a typo there, i meant cotangent, not cosine

Answer 32

yes, tangent is the reciprocal of cotangent

Answer 33

let me tell you what you should know besides " SOA CAH TOA"

Answer 34

you should know that \[\tan(x)=\frac{\sin(x)}{\cos(x)}\] and \[\csc(x)=\frac{1}{\sin(x)}\] and \[\sec(x)=\frac{1}{\cos(x)}\]

Answer 35

Without having to think

Answer 36

using those identities, and \[\cos^2(x)+\sin^2(x)=1\] and algebra, you can solve most trig identities

Answer 37

yes, without thinking
just know them

Answer 38

ready to tackle the last one?

Answer 39

I need to put them on index cards.

Answer 40

Review them daily^

Answer 41

Yes I am. If I reply slow, it's ebcause OS is running slow for me now.

Answer 42

i just sent you a decent cheat sheet, but yes, you should write them down yourself so you will know them

Answer 43

yeah slow for me too, no problem

Answer 44

Thank you for the sheet!

Answer 45

yw
let me know when you are ready for the last one

Answer 46

\[\csc^{2}x-1=\cot^{2}x\]

Answer 47

ok math teacher love this stuff
this is where you start, with the mother of all identities,the first one \[\cos^2(x)+\sin^2(x)=1\]

Answer 48

you don't have that, but you do have cosecant and cotangent
do you remember what cosecant is? i wrote it above

Answer 49

you mean H/O or recip of sin?

Answer 50

second one
the reciprocal of sine

Answer 51

Okay.

Answer 52

so starting with \[\cos^2(x)+\sin^2(x)=1\] how to you get \(\csc(x)\)?
answer: divide everything by \(\sin(x)\) that way \(\csc(x)\) will turn up on the right

Answer 53

so it would be cos(x)/sin(x)

Answer 54

Or am I wrong

Answer 55

\[\cos^2(x)+\sin^2(x)=1\\
\frac{\cos^2(x)}{\sin^2(x)}+\frac{\sin^2(x)}{\sin^2(x)}=\frac{1}{\sin^2(x)}\]

Answer 56

i made a mistake, i should have said divide all by \(\sin^2(x)\)

Answer 57

now you have almost exactly what you want

Answer 58

\[\frac{1}{\sin^2(x)}=\csc^2(x)\] on the right

Answer 59

evidently \[\frac{\sin^2(x)}{\sin^2(x)}=1\]

Answer 60

that comes from csc=1/sin correct? for the above? jsut to make sure

Answer 61

yes
what about \[\frac{\cos^2(x)}{\sin^2(x)}\] what is that ?

Answer 62

well.. cos/sin = cot, right? so cot^2(x)?

Answer 63

zactly

Answer 64

Wow. I actually got that..

Answer 65

so you have \[\cot^2(x)+1=\csc^2(x)\]

Answer 66

yes, you did !

Answer 67

With your expert guidance of course.

Answer 68

Can you think of any values that x can not have?

Answer 69

you get from \[\cot^2(x)+1=\csc^2(x)\]to \[\csc^{2}x-1=\cot^{2}x\] using elementary algebra

Answer 70

Okay.

Answer 71

aka add 1 to both sides

Answer 72

ok not really, subtract 1 from both sides

Answer 73

I was gonna say you subtracted.

Answer 74

Dont lose me now! xD

Answer 75

lol we are done in any case

Answer 76

now i have a question:
what the monkey is a "turkey committee"?

Answer 77

Haha, I think you have asked me before, I am almost positive. xD I am in 4-H and we show poultry, turkeys being one of the birds. There is committees that help run and function each species, including turkeys. I am part of the committee that helps assist other kids showing the project, taking care, being the advisor, etc.

Answer 78

I basically am like, co-boss.

Answer 79

actually you are right , i remember now

Answer 80

ok good luck with the trig, i am off to watch perry mason

Answer 81

Haha, okay, thank you so much for your help. You are my hero tonight.

Answer 82

yw, my pleasure to help someone who is willing to help themselves

Answer 83

:) Take care!