Question

@callisto Help please?
I am going to answer a trig id question by myself. Can you just check my procedure?
Thank you

Question: verify the pic.

Answer 1

Oh... This is troublesome. This time, you can do both side. BTW, I owed you the answer of your last question yesterday :S

Answer 2

I already answered:) Can you check this?

Answer 3

I remember the question, don't worry :)

Answer 4

Wow, I wish I had your memory:)

Answer 5

my memory is like a sieve lol
Back to the question, i did it more or less the same as you've done. :)
Here's my workings:
RHS = (1-cos2x)/(1+cos2x)
= [ 1- (1-2sin²x)]/[ 1+ (2cos²x-1)]
= 2sin² x / 2cos²x
= sin² x / cos² x
= tan²x
= LHS

Answer 6

So my answer is ok right?

Answer 7

Yup :)

Answer 8

Great:) Ok so for this question, I have worked on it abit, can you check my procedure?

Answer 9

Okay :)

Answer 10

Ok:)
So first LHS:

sin² x - cos² x - tan² x
=sin² x - cos² x - (sinx / cosx)²
sin²(45) - cos²(45) - (sin(45)/cos(45))²
= (1/sqrt(2))²)- (1/sqrt(2))²)-( (1/sqrt(2))²)/ (1/sqrt(2))²))

Answer 11

After, that Im not sure wether to factor out everything and get 0, or to solve and get -1

Answer 12

Solve it , I think :P
but you'll get -1/4

Answer 13

Am i doing it right though? Im following book examples and exercises, so im not even sure Im going the right way

Answer 14

And if I solve, dont I get this?

http://www.wolframalpha.com/input/?i=%281%2Fsqrt%282%29%29%C2%B2+-+%281%2Fsqrt%282%29%29%C2%B2+-+%28%281%2Fsqrt%282%29%29%C2%B2+%2F+%281%2Fsqrt%282%29%29%C2%B2%29#

Answer 15

ooppsss... I was wrong... neglect me :(

Answer 16

Oh. So -1 is ok?

Answer 17

Im solving it correctly right?

Answer 18

sin² x - cos² x - tan² x
sin²(45) - cos²(45) - (tan²(45))
= (1/sqrt(2))²)- (1/sqrt(2))²)-( 1²)
=-1
You don't need to convert tanx to sinx/cosx in this case. Though, you got the right answer :)

Answer 19

Awesome:) so now that we determined that LHS is -1. I will show you whwat Ive done so far in RHS

Answer 20

[2sin² x - 2sin^4 x - 1] / [1-sin² x]
= [2sin² (45) - 2sin^4 (45) - 1] / [1-sin² (45)]
=[2*(1/sqrt(2))² - 2*(1/sqrt(2))4 - 1] / 1 - (1/sqrt(2))²
=If im not wrong here, we factor out the 1s and the (1/sqrt(2))²
so we will be left with 2*2 (1/sqrt(2))^4

Answer 21

Which will result in 1

Answer 22

If what I did is correct, then it implies that they are not identities right?

Answer 23

Or did i so something wrong in my answer?

Answer 24

OHH I did do something wrong

Answer 25

its not 4, its -4

Answer 26

which will give me -1 as an answer

Answer 27

because 2* -2 = -4.
therefore,
it is -4* (1/sqrt(2))^4 = -1

Answer 28

So they are identities, if what I did is correct, then I proved that they are identities

Answer 29

Did i solve it correctly?

Answer 30

Hmm, they are equal.
[2sin² x - 2sin^4 x - 1] / [1-sin² x]
=[2sin² x - 2sin^4 x - 1] / [cos² x]
= [2(1/2) - 2(1/4) -1]/ (1/2)
= (-1/2) / (1/2)
=-1
:)

Answer 31

Wait, did I do it wrong in my procedure? Because we got the same asnwer, but it seems you did something differently.

Answer 32

lol, I tried to do it as simple as possible :P

Answer 33

Oh, But is my procedure correct?

Answer 34

Just a minute

Answer 35

Kae:)

Answer 36

=[2*(1/sqrt(2))² - 2*(1/sqrt(2))^4 - 1] / 1 - (1/sqrt(2))²
= (1-1/2-1)/ (1/2)
= you'll get the same answer as mine :)

Answer 37

So it is ok right? one question,in yours why did put 1/2 when sin 45 = (1/sqrt(2))?

Answer 38

becausee it's squared :P

Answer 39

ohh ok:) So does my procedure make sense? Do you think its good to submit?

Answer 40

It's okay :)
But double check your answer. especially the sqrt things !

Answer 41

Ok:), Ill do that, Thank you Jos:)

Answer 42

welcome :)