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

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

Mathematics
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review some basics |dw:1432906516081:dw|
That's a good graph. I get that for the most part...but I have no idea how to verify identities
for example csc(theta) - sin(theta) = cot(theta) cos(theta) I have to verify the identity and find the domain

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@B_O_R_E_D Are you familiar with the basic identities?
sort of-ish...they are confusing
Can you post what you know? We'll see if you have everything you need.
hang on
1 Attachment
Sorry, I was expecting you to type what YOU know, not what's online.
its not online its just a word doc...I can type if youd like
Sorry, I was expecting you to type what YOU know, not what's online. Besides, I only open/display .jpg files.
okay...hang on I'll type it
Yes, it would be nice. Because typing them out also helps you getting familiar with them.
|dw:1432907494538:dw|
@triciaal what you posted is a very good review of the basics, before even the question was posted. Genial!
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)
reciprocal identities
Did you miss out anything? There is at least one more that you need.
tan(theta) = sin(theta)/cos(theta) (Tan identity) cot(theta) = cos(theta)/sin(theta) (Cot Identity)
Good, we'll need these and... one more...
that's all I've learned
|dw:1432904305976:dw| This is extracted from @triciaal 's post. Can you use Pythagoras Theorem to write down the missing identity?
ummm sin(theta)/tan(theta)?
Think Pythagoras's theorem!
SOHCAHTOA
???
|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?
Hint: use the diagram I extracted from @Triciaal 's post.
hang on
I really have no idea... ummm did I miss cos?
more hint: |dw:1432904684712:dw| Pythagoras theorem says \(a^2+b^2=c^2\)
yes I know that
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?
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
But now you understand why this is true?
yeah more or less
im leaning toward less
okay...so I know we have to use a reciprocal identity
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.
okay....
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.
If we can show that the left side is identical to the right, we're done.
okay.. so I think I need to get like similar terms or something so would I replace sin(theta) for 1/csc(theta) ???
Exactly, that's an excellent start. A lot of the time, this strategy works!!!
yay!! okay so csc(theta) - 1/csc(theta) = cot(theta) cos(theta)
I don't see your reasoning, can you explain?
I replaced sin...so I was just filling that in the equation
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)
Are you at ease with adding fractions, such as 1/3 + 3?
oh! I put it on the wrong side yeah those are fine
1/3-3 = 1/3 - 3^2/3 = (1-3^2)/3 Does this make sense to you?
okay.... oh yes! I see what you did okay
Now you need to do the same with 1/sin(x) and continue the following \( csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)\)
umm okay...
R U stuck or R U working on it?
trying to work on it...but im still stuck
Well... it's similar to the 1/3 example: \(\large csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)=\frac{1-sin^2(x)}{sin(x)}\)
okay....so that's how you verify it?
or \(\large csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)=\frac{1}{sin(x)}-\frac{sin(x)}{sin(x)}sin(x)=\frac{1-sin^2(x)}{sin(x)}\)
With what you know in terms of trigonometric identities, you should be able to finish the job, can you not?
csc(x) - sin(x) = 1?
Anyway, from the identity \(sin^2(x)+cos^2(x)=1\) we conclude: \(1-sin^2(x)=cos^2(x)\)
Does that help?
ohhhh...oops sorry... OHHH!!! I get it! yes that helps a lot!
Thank you so much for your help/time!!
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.
okay thank you. I think I can do that... I have to go, but thank you again!
You're welcome! :)
sorry system was down @mathmate thank you

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