I desperately need someone to help me with Trig!!! please!!! FAN AND MEDAL!!

- anonymous

I desperately need someone to help me with Trig!!! please!!! FAN AND MEDAL!!

- Stacey Warren - Expert brainly.com

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- katieb

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- triciaal

review some basics
|dw:1432906516081:dw|

- anonymous

That's a good graph. I get that for the most part...but I have no idea how to verify identities

- anonymous

for example csc(theta) - sin(theta) = cot(theta) cos(theta)
I have to verify the identity and find the domain

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## More answers

- mathmate

@B_O_R_E_D Are you familiar with the basic identities?

- anonymous

sort of-ish...they are confusing

- mathmate

Can you post what you know? We'll see if you have everything you need.

- anonymous

hang on

- anonymous

##### 1 Attachment

- mathmate

Sorry, I was expecting you to type what YOU know, not what's online.

- anonymous

its not online its just a word doc...I can type if youd like

- mathmate

Sorry, I was expecting you to type what YOU know, not what's online.
Besides, I only open/display .jpg files.

- anonymous

okay...hang on I'll type it

- mathmate

Yes, it would be nice. Because typing them out also helps you getting familiar with them.

- triciaal

|dw:1432907494538:dw|

- mathmate

@triciaal what you posted is a very good review of the basics, before even the question was posted. Genial!

- anonymous

csc(theta) = 1/sin(theta)
sin(theta) = 1/csc(theta)
sec(theta) = 1/cos(theta)
cos(theta) = 1/sec(theta)
tan(theta) = 1/cot(theta)
cot(theta) = 1/tan(theta)

- anonymous

reciprocal identities

- mathmate

Did you miss out anything?
There is at least one more that you need.

- anonymous

tan(theta) = sin(theta)/cos(theta) (Tan identity)
cot(theta) = cos(theta)/sin(theta) (Cot Identity)

- mathmate

Good, we'll need these
and... one more...

- anonymous

that's all I've learned

- mathmate

|dw:1432904305976:dw|
This is extracted from @triciaal 's post.
Can you use Pythagoras Theorem to write down the missing identity?

- anonymous

ummm sin(theta)/tan(theta)?

- mathmate

Think Pythagoras's theorem!

- anonymous

SOHCAHTOA

- anonymous

???

- mathmate

|dw:1432904481376:dw|
Perhaps I missed out the 90 deg. angle didn't help.
Pythagoras theorem says that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse.
Can you figure out something?

- mathmate

Hint: use the diagram I extracted from @Triciaal 's post.

- anonymous

hang on

- anonymous

I really have no idea... ummm did I miss cos?

- mathmate

more hint:
|dw:1432904684712:dw|
Pythagoras theorem says \(a^2+b^2=c^2\)

- anonymous

yes I know that

- mathmate

ok, use the diagram below and use Pythagoras theorem:
|dw:1432904785881:dw|
It would be
\(sin^2(x)+cos^2(x) = 1\)
This is the missing identity.
Have you seen that before?

- anonymous

ohhhhh
no I haven't....They have just introduced trig to me in this last unit for the year...everything I've done this year has been alg2

- mathmate

But now you understand why this is true?

- anonymous

yeah more or less

- anonymous

im leaning toward less

- anonymous

okay...so I know we have to use a reciprocal identity

- mathmate

The identity comes from the definition of sin and cos. With a hypotenuse of 1, the identity is just an application of Pythagoras.
In any case, you only have to know it, most of the time you're not expected to prove it.

- anonymous

okay....

- mathmate

Now back to the question.
csc(theta) - sin(theta) = cot(theta) cos(theta)
In verifying identities, we usually start from one side.
In this case, we can start from the left.

- mathmate

If we can show that the left side is identical to the right, we're done.

- anonymous

okay.. so I think I need to get like similar terms or something so would I replace sin(theta) for 1/csc(theta) ???

- mathmate

Exactly, that's an excellent start.
A lot of the time, this strategy works!!!

- anonymous

yay!!
okay so
csc(theta) - 1/csc(theta) = cot(theta) cos(theta)

- mathmate

I don't see your reasoning, can you explain?

- anonymous

I replaced sin...so I was just filling that in the equation

- mathmate

If we start with the left, we don't usually write the right hand side, because equality is not yet proven.
We write (I'll use x instead of \(\theta\), saves me time.) as your first step:
csc(x)-sin(x) = 1/sin(x) - sin(x)

- mathmate

Are you at ease with adding fractions, such as 1/3 + 3?

- anonymous

oh! I put it on the wrong side
yeah those are fine

- mathmate

1/3-3 = 1/3 - 3^2/3 = (1-3^2)/3
Does this make sense to you?

- anonymous

okay....
oh yes! I see what you did okay

- mathmate

Now you need to do the same with 1/sin(x) and continue the following
\( csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)\)

- anonymous

umm okay...

- mathmate

R U stuck or R U working on it?

- anonymous

trying to work on it...but im still stuck

- mathmate

Well... it's similar to the 1/3 example:
\(\large csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)=\frac{1-sin^2(x)}{sin(x)}\)

- anonymous

okay....so that's how you verify it?

- mathmate

or
\(\large csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)=\frac{1}{sin(x)}-\frac{sin(x)}{sin(x)}sin(x)=\frac{1-sin^2(x)}{sin(x)}\)

- mathmate

With what you know in terms of trigonometric identities, you should be able to finish the job, can you not?

- anonymous

csc(x) - sin(x) = 1?

- mathmate

Anyway, from the identity
\(sin^2(x)+cos^2(x)=1\)
we conclude:
\(1-sin^2(x)=cos^2(x)\)

- mathmate

Does that help?

- anonymous

ohhhh...oops sorry... OHHH!!! I get it! yes that helps a lot!

- anonymous

Thank you so much for your help/time!!

- mathmate

Actually, looking back, you have been asked to "verify" the identity.
What we did was to prove it.
Verification means putting numbers on each side and show that the results are the same.
You need to put different numbers, but make sure theta does not equal zero (because the expression becomes infinite, that's where the domain comes in).
Sorry for having gone through long steps, but this will also prepare for your future topic on identities.

- anonymous

okay thank you. I think I can do that...
I have to go, but thank you again!

- mathmate

You're welcome! :)

- triciaal

sorry system was down
@mathmate thank you

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