Question about simplifying this equation...

iBankingFTW

tan x/sin x + 1/cos x

The choices are A. 2tan^2x B. 2cosx C. 2cosx-1 D. 2secx

Do I make the tan sin/cos and multiply by reciprocal of sin? I did that and end up with 1/cos + 1/cos

mathman

You got it right so far. Do you know what secx is?

Diffy

Tan x / Sin x + 1 /cos x

Since tanx = (sinx/cosx) we can substitute and get:

(Sinx / cos x)/ Sin x + 1 / cos x

here we can cancel out the sin x 's and get

1 / cos x + 1 / cos x

since the denominators are the same we can add and get

2 / cos x

Now since sec x = 1 / cosx we get

2 sec x


Make sense?

iBankingFTW

Yeah, thanks for that. But I thought about what mathman said and the answer was obvious, lol. Thanks a lot though for working it out.

Yeah, so the 2(1/cosx) becomes 2secx...

That was pretty stupid, lol. Thanks though. This is a little off-topic for this thread but I really lack in courses like physics and math and I really like this forum (I'm new, lol) and what I like is how you guys don't just give an answer. You either show all the steps like Diffy did or what you did and made me think about it. :)

iBankingFTW

I have another question I can't understand.

Which expression is equivalent to cos(pi + theta) (I don't know how to import the signs. and I checked the computer's character map)

So what I did was take cos(pi + theta) and put down the formula:

cos pi * cos theta - sin pi * sin theta

But I don't know what to do from there. I tried to see if I could do anything putting sin over cos but they're on opposite sides of the "-".

Choices are: A. -cos theta B. cos theta C. -sin theta D. sin theta

Diffy

you have cos(pi) and Sin(pi) in your expressions... What is the cos(pi)? sin(pi)?

Diffy

Oh and for learning how to import the symbols, you can use Latex tags, there is a guide on this site somewhere, [tex]\pi\theta[/tex]

Edit:

Here is the link to the Latex tutorial:
http://www.physicsforums.com/showthread.php?t=8997

iBankingFTW

Thanks for telling me about the symbols. That should come in handy :)


Just realized this isn't the forum for homework questions so I'm going to make a new thread in the correct section.