Question

Verify the following.

1. 1-cot^2 x sin^2 x = sin^2 x

NOTE: x= the zero degree sign I

Answer 1

I wasn't able to write the correct sign on here so i used x instead

Answer 2

theta?

Answer 3

yes that sign thank you

Answer 4

cotan and sin are multiplied?

Answer 5

yes

Answer 6

\[1-\cot ^{2} \theta \sin ^{2} \theta = \sin ^{2}\theta\]

Answer 7

^ yes thats how the problem looks

Answer 8

maybe divide through by sin^2 theta

Answer 9

which will give you 1/ sin ^2 x - cot^2 x on the left and 1 on the right

Answer 10

cot = 1/tan, tan = sin/cos, so cot = cos/tan

Answer 11

sorry not cos/tan, cos/sin

Answer 12

I think the problem might want you to use a trig identity to prove \[\sin ^{2}\]

Answer 13

yes it wants me to use the identities

Answer 14

\[\sin ^{2}\theta + \cos ^{2}\theta = 1\]

Answer 15

yes thats the one.

Answer 16

so multiply through by sin^2 x. then change cot^2 x into (cos^2 x)/(sin^2 x) so you now have...

Answer 17

\[1 + \cot ^{2}\theta = \csc ^{2}\theta \]

Answer 18

\[\frac{1 - \cos^{2} \theta} {\sin^2 \theta} = 1\]

Answer 19

then just multiply by sin^2 theta and you have the identity

Answer 20

now back to call of duty :D

Answer 21

OK thank you both so much now i just have to follow through to make sure that i understand it

Answer 22

try and do the steps i did, and tell me if you get stuck

Answer 23

instead of multiplying sin though she needs to divide

Answer 24

your identity is equal to sin not cosecant

Answer 25

yes thats right, at the start divide by sin^2

Answer 26

oh ok thank you

Answer 27

done it?

Answer 28

still working on it give me a minute

Answer 29

no prob

Answer 30

ok in stuck and confused

Answer 31

show me where your up to

Answer 32

you have to verify that the equation equals sin^2 theta

Answer 33

your trying to prove that the left side = the right side.

Answer 34

so you would just look at the problem being 1-cot^2thetasin^2theta

Answer 35

so if you can get the equation into the form \[1 = \cos^{2} + \sin^{2}\] Which we know is always true. then you have proved it

Answer 36

your trying to manipulate it into that form

Answer 37

or one of these http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 38

here were trying to get it into the "pythagorean identity"

Answer 39

yes

Answer 40

okay so youve divided by sin^2 x

Answer 41

\[\frac{1}{\sin^{2} x} - \cot^{2} x = 1\] and you got that?

Answer 42

wait when i divide sin^2 x i divide it on both sides correct?

Answer 43

then using the identity that \[\cot^2 x = \frac{1}{\tan^2 x} = \frac{\cos^2 x}{\sin^2 x}\]

Answer 44

yeah on both sides, to keep it balanced

Answer 45

exactly

Answer 46

but if i divide it on both sides i would get 1-cot^2x=1 i dont get how you got something different

Answer 47

okay so because \[\cot^2 x = \frac{1}{\tan^2 x} = \frac{\cos^2 x}{\sin^2 x}\] we can now rewrite it as...
\[\frac{1}{\sin^{2} x} - \frac{\cos^2 x}{\sin^2 x}\ = 1\]

so we replaced the cot term with the cos/sin, is there anything you dont get?

Answer 48

no you have to divide the 1 at the front by sin^2 too

Answer 49

got to do the same thing to every term

Answer 50

to keep it nice and balanced

Answer 51

do you know that sin = opposite / hypotenuse, cos = adjacent/ hypotenuse, tan = opposite/adjacent ?

Answer 52

yes now i understand how you got to that point

Answer 53

so now that i get to the point 1/sin^2x-cos^2x/sin^2x can i join them together and write it as 1-cos^2x/sin^2x which will then equal sin^2/sin^2= 1 ?

Answer 54

you need to use parenthesis

Answer 55

no keep it like this \[\frac{1}{\sin^{2} x} - \cot^{2} x = 1\] have you got to that stage?

Answer 56

what you should have jusging by what you've written it should be csc^2(x)-cot^2(x)

Answer 57

im solving it for sin^2 x + cos^2 x = 1

Answer 58

and simplify jame's equation and you have your identity

Answer 59

you take over veritas

Answer 60

\[\csc ^{2}\theta + \cot ^{2}\theta= 1 \rightarrow 1- \cot ^{2}\theta = \csc ^{2}\theta \]

Answer 61

ok i am very confused right now so now that i got to 1/sin^2 x - cot^2x

Answer 62

they dont have the same denominator so do i still change the cot^2 x into the csc^2 x i am very confused with that part

Answer 63

wait no now i get it

Answer 64

you changed the 1/sin^2 into the csc right?

Answer 65

yes he did

Answer 66

exactly that, have you solved it now? :)

Answer 67

no i just got csc^2x-cot^2x=1
do i add the cot^2x over to the other side?

Answer 68

yeah cot to the other side and you have nailed it

Answer 69

ok thank you both for your patience and for your help i really appreciate it

Answer 70

no problem, glad you understand it

Answer 71

Me too lol

Answer 72

Just anoying that it doesnt get much easier. Its a constant struggle to grasp the next thing im learning :(

Answer 73

i definitely understand where your coming from ='( lol