Oye, the dreaded trig identities

[K.ourt]

Yup, our final is coming up and our teacher told us that one of the four identities she gave us would be on it, so I want to make sure I have these right.

I think it's right, and it checks...so yay?

I can't seem to get it any farther without it not checking out. I cancelled out the sine's, leaving me with (1-sinx)/(cosxsinx) But it didnt check in my calculator...oh, and those are supposed to be theta signs, even though half of them don't look like it. Heh.

 

[tkhunny]

I cancelled out the sine's, leaving me with (1-sinx)/(cosxsinx) But it didnt check in my calculator

Small wonder. What does that even mean? Cancelled what sines? It looks to me like you magically discarded the exponent. What was that supposed to do? Don't do that.

(1-sin^2(x))/(cos(x)sin(x))

1-sin^2(x) = cos^2(x)

 

[Guest]

The first one doesn't look right to me. Doesn't (sinx/cosx)(sinx/cosx)=(sin^(2)x)(cos^(2)x)?

Edit.. Ignore the rest of what I had here. It was definately wrong.

 

[tkhunny]

On the first. What checks? I can;t find anythign that "checks". It is time for you to review basic algebra. You changed tan(x) and cot(x) to equivalent expressions for sine and cosine. That was the last thing you did correctly.

Do you really believe:

(1+2)/(1+3) = 2/3 ??? Prove by cancelling the '1's.

Let's see

(1+2)/(1+3) = (3)/(4) = 3/4 ≠ 2/3 I guess that '1'-cancelling stuff doesn't work. Never do that again.

How does sin(x)*sin(x) = 2*sin(x)? Very basic algebra. Did you skip a class?

I'll show you all of the first one. If you cannot follow this, without additional commentary, you are in the wrong class. Go talk to your teacher immediately about appropriate placement.

(1+tan(x))/(1+cot(x)) =

(1+(sin(x)/cos(x)))/(1+(cos(x)/sin(x))) =

((cos(x)/cos(x))+(sin(x)/cos(x)))/((sin(x)/sin(x))+(cos(x)/sin(x))) =

((cos(x) + sin(x))/cos(x))/((sin(x) + cos(x))/sin(x)) =

[(cos(x) + sin(x))*sin(x)]/[cos(x)*(sin(x) + cos(x))] =

[(cos(x) + sin(x))/(sin(x) + cos(x))]*[sin(x)/cos(x)] =

[1]*[sin(x)/cos(x)] =

[sin(x)/cos(x)] = tan(x) at last.

Good luck.

 

[K.ourt]

Thank you, and I am well aware that I am not up to level for PM30. The only problem with that is that I need it to get into college (for Social Work, none-the-less) and I have a mother who doesnt understand breathing down my neck. I assure you that I would not be butchering it if I had a choice in the matter. But thank you for the 'tough love' and help, I know it must be painful for you to have to decipher through all the wrong.

 

[tkhunny]

Well, you do have to get a few things straight in order to get good answers. Sometimes you can just run a little test to see if things actually work the way you have implemented them.

Casting out 1's is a bit odd, and its invalidity can be demonstrated as I did above.

Good luck, keep trying, and try to remember that I'm really not hostile. I know it sounds like it most of the time.