Question

Proving trigonometric identities.
Prove that:
((sin(x)-cos(x)-1)/(sin(x)+cos(x)-1))=((cos(x)+1)/(sin(x))

Answer 1

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

Answer 2

I suggest we cross multiply then simplify both sides until we get to a true identity.

Answer 3

Begin with the more complicated side and simplify until you obtain the expression on the other side. Which side do you think you should start with @blackbird02 ?

Answer 4

i think we should start with the left side.

Answer 5

Cross multiplication is a cheap way out for a lack of a better phrase.

Answer 6

Yep. Yep.

Answer 7

Yes! @blackbird02 start with the left side.

Answer 8

This will not go well if you only work on one side, we work on both sides
simultaneously until we get to a true identity.

Answer 9

should i do this?
\[\frac{ \frac{ 1 }{ \csc(x) }-\frac{ 1 }{ \sec(x) }-1 }{ \frac{ 1 }{ \csc(x) }+\frac{ 1 }{ \sec(x) }-1}=\frac{ \cos(x)-1 }{ \sin(x) }\]

Answer 10

no, it will not help

Answer 11

Not necessary! No

Answer 12

I was suggesting you to cross multiply... @blackbird02

Answer 13

How do I work on the left side? What should I do first?

Answer 14

@Miracrown Besides cross multiplying, what could be other steps? We're actually not allowed to cross multiply.

Answer 15

I was saying, we cannot work only on one side
we need to work on BOTH sides at the same time.... cross multiply, then simplify.

Answer 16

whatever you do, you will do a form of cross multiplicaiton for this one

Answer 17

Okay well I see you prefer @Miracrown to help you. Good luck. Bye.

Answer 18

Hey @calculusfunctions, don't be too hasty. =]

Answer 19

@calculusfunctions Well, I could use a little advice from you too. What do you think?

Answer 20

I do not think there is any other way except cross multiplication.
if you are not allowed to do that,
I do not think I am able to solve it.

Adios!

Answer 21

Other ways I can think is to transform them to other functions like what I posted above, but I'm stuck from there.

Answer 22

@Miracrown I wasn't being hasty. I just didn't want to waste my time if you're already helping.

Answer 23

Are you tricking us @blackbird02? This is not an identity that is why nothing works, the left side is NOT equal to the right side.

Answer 24

First @blackbird02 I think you copied the problem incorrectly. The right side should be\[-\frac{ \cos x +1 }{ \sin x }\]

Answer 25

@blackbird02 Sorry, may be the right side equal to this?
\[-\frac{ 1+\cos(x) }{ \sin(x) }\]

Answer 26

Right @blackbird02 that's what I just said. Now it works.

Answer 27

@calculusfunctions yes, that's right.

Answer 28

@blackbird02 I'll get you started. First multiply the numerator and the denominator of the left side by sinx + cosx +1. Go ahead and post the result of this multiplication and then I'll walk you through the rest

Answer 29

Make sure you multiply both the numerator and the denominator by\[\sin x +\cos x +1\]

Answer 30

Ok, just a sec

Answer 31

@Michele_Laino I hope you're not posting a solution! He needs to learn NOT look at someone else's work. Thank you!

Answer 32

Good advice - @calculusfunctions.

Answer 33

@calculusfunctions Ok!, done!

Answer 34

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

Answer 35

is this correct?

Answer 36

Excellent @blackbird02 Now the next step??

Answer 37

Do recognize the Pythagorean Identity in the denominator @blackbird02 ?

Answer 38

It's sin^2A+cos^2B = 1, right?

Answer 39

Yes!

Answer 40

Also write\[\sin ^{2}x\]in the numerator in terms of\[\cos ^{2}x\]using the same Pythagorean Identity of course.

Go Ahead!!

Answer 41

@blackbird02 ??

Answer 42

@calculusfunctions I'm working on it

Answer 43

Okay no problem! Just wanted to make sure you're still there. Take your time.

Answer 44

I'm not really sure if I did this correctly. Can you check it?

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

Answer 45

No! From your first step which was correct, you had\[\frac{ \sin ^{2}x -\cos ^{2}x -2\cos x -1}{ \sin ^{2}x +2\sin x \cos x +\cos ^{2}x -1 }\]If you replace\[\sin ^{2}x +\cos ^{2}x =1\]in the denominator, then what should that denominator be now?

Answer 46

The denominator would now be \[2\sin(x)\cos(x)\] ?

Answer 47

Yes!

Answer 48

And if you write \[\sin ^{2}x\]in the numerator in terms of cosine, then what should the numerator be?

Answer 49

The whole fraction would be:
\[\frac{ 1-\cos^2(x)-\cos^2(x)-2\cos(x)-1 }{ 2\sin(x)\cos(x) }\]

Answer 50

Very Good @blackbird02 !! Now collect like terms in the numerator.

Answer 51

Simplifying it would be:
\[\frac{ -2\cos^2(x)-2\cos(x) }{ 2\sin(x)\cos(x) }\]

Answer 52

What do I do next?

Answer 53

Excellent! Now do you recognize greatest common factor in the numerator?

Answer 54

cos(x) ? I'm not really sure.

Answer 55

What is the greatest common factor in the numerator? Do you not see that -2cosx is common to both terms in the numerator?

Answer 56

So, I'll factor it out?

Answer 57

Yes! Go ahead!

Answer 58

Ok, I got it now! @calculusfunctions thank you so much for helping me out and guiding me throughout the process. I really appreciate it.

Answer 59

No Problem!! Good Luck with everything!