trigonometric identities

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September 14th 2008, 09:43 AM
robasc

trigonometric identities

I am trying to learn how to transform identities of one side of the equation to the other side. is this correct?

I am using 0 for theta
theta = 0

question:

tan 0 + cot 0
-------------- = csc^2 0

tan 0

answer:

cot 0 = csc 0
September 14th 2008, 09:50 AM
Jhevon

Quote:

Originally Posted by robasc

I am trying to learn how to transform identities of one side of the equation to the other side. is this correct?

I am using 0 for theta
theta = 0

question:

tan 0 + cot 0
-------------- = csc^2 0

tan 0

answer:

cot 0 = csc 0

that answer is wrong. how does that prove an identity. you must show the left hand side and right hand side are the same. so you must turn one side into the other or bring both sides to the same thing

consider the left hand side (it is more complicated and so gives us more options to change things)

note, $\frac {\tan \theta + \cot \theta}{\tan \theta} = \frac {\tan \theta}{\tan \theta} + \frac {\cot \theta}{\tan \theta}$

now what?
September 14th 2008, 10:02 AM
robasc

On the left side I know that I will be left with tan 0.

but the right side confuses me.

wait I think I got it:

tan 0 = sin 0 / cos 0
September 14th 2008, 10:06 AM
Jhevon

Quote:

Originally Posted by robasc

On the left side I know that I will be left with tan 0.

but the right side confuses me.

wait I think I got it:

tan 0 = sin 0 / cos 0

yes, it you can use that

now how do we finish the problem?
September 14th 2008, 10:16 AM
robasc

so the answer is tan 0?
September 14th 2008, 10:20 AM
Jhevon

Quote:

Originally Posted by robasc

so the answer is tan 0?

huh??

lets not lose sight of the objective here. we are proving an identity. we are trying to turn one side into the other. your objective is to turn $\frac {\tan \theta}{\tan \theta} + \frac {\cot \theta}{\tan \theta}$ into $\csc^2 \theta$ through simplification. that will show that the left hand side and right hand side are the same
September 14th 2008, 11:18 AM
11rdc11

I'm going to work this out for you since it seems you are a little lost and hopefully me working it will help you understand, but I what you to come back and explain the steps I did so you will learn.

$\frac {\tan \theta + \cot \theta}{\tan \theta}$

$=~\frac {\tan \theta}{\tan \theta} + \frac {\cot \theta}{\tan \theta}$

$= 1+ \frac{cos^2\theta}{sin^2\theta}$

$= \frac{sin^2\theta}{sin^2\theta} + \frac{cos^2\theta}{sin^2\theta}$

$=~\frac{1}{sin^2\theta}$

$=~csc^2\theta$
September 14th 2008, 11:20 AM
robasc

ok I have come up with this:

sin/cos + cos/sin
------------------
sin/cos + sin/cos

cos/sin
-------
sin/cos

cos/sin * cos/sin =

cos^2 0
--------
sin^2 0

this is how I am thinking about it?
September 14th 2008, 11:27 AM
Jhevon

Quote:

Originally Posted by robasc

ok I have come up with this:

sin/cos + cos/sin
------------------
sin/cos + sin/cos

evidently you need help with adding fractions (see posts #3, 9 and 11 here). you cannot cancel like that! and you cannot have sin and cos by themselves, write sin(x) and cos(x) or something

furthermore, as i said, you are proving an identity. if you don't end up with something that looks like the other side, you are wrong

Quote:

this is how I am thinking about it?

you're asking us or telling us? :p

11rdc11 did the problem for you. make sure you follow what he did
September 14th 2008, 11:36 AM
robasc

I was telling you how I was interpreting the problem.
September 14th 2008, 11:38 AM
11rdc11

Oops I didn't notice that help canceled out $\frac{sin\theta}{cos\theta}$ like that
September 14th 2008, 11:49 AM
Jhevon

Quote:

Originally Posted by robasc

I was telling you how I was interpreting the problem.

yes, i know. i was messing with you because you put a question mark at the end.

study what 11rdc11 did and make sure you have no questions. then, make sure you review fractions and trig identities. things like $\sin^2 \theta + \cos^2 \theta = 1$ , stuff like that you should know by heart. you are in for a LOT of pain if you do not get these basic things. you don't want to be one of those students that end up in calculus and can't do basic algebra, trust me