anonymous
  • anonymous
I desperately need someone to help me with Trig!!! please!!! FAN AND MEDAL!!
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
triciaal
  • triciaal
review some basics |dw:1432906516081:dw|
anonymous
  • anonymous
That's a good graph. I get that for the most part...but I have no idea how to verify identities
anonymous
  • 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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mathmate
  • mathmate
@B_O_R_E_D Are you familiar with the basic identities?
anonymous
  • anonymous
sort of-ish...they are confusing
mathmate
  • mathmate
Can you post what you know? We'll see if you have everything you need.
anonymous
  • anonymous
hang on
anonymous
  • anonymous
1 Attachment
mathmate
  • mathmate
Sorry, I was expecting you to type what YOU know, not what's online.
anonymous
  • anonymous
its not online its just a word doc...I can type if youd like
mathmate
  • mathmate
Sorry, I was expecting you to type what YOU know, not what's online. Besides, I only open/display .jpg files.
anonymous
  • anonymous
okay...hang on I'll type it
mathmate
  • mathmate
Yes, it would be nice. Because typing them out also helps you getting familiar with them.
triciaal
  • triciaal
|dw:1432907494538:dw|
mathmate
  • mathmate
@triciaal what you posted is a very good review of the basics, before even the question was posted. Genial!
anonymous
  • 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
  • anonymous
reciprocal identities
mathmate
  • mathmate
Did you miss out anything? There is at least one more that you need.
anonymous
  • anonymous
tan(theta) = sin(theta)/cos(theta) (Tan identity) cot(theta) = cos(theta)/sin(theta) (Cot Identity)
mathmate
  • mathmate
Good, we'll need these and... one more...
anonymous
  • anonymous
that's all I've learned
mathmate
  • mathmate
|dw:1432904305976:dw| This is extracted from @triciaal 's post. Can you use Pythagoras Theorem to write down the missing identity?
anonymous
  • anonymous
ummm sin(theta)/tan(theta)?
mathmate
  • mathmate
Think Pythagoras's theorem!
anonymous
  • anonymous
SOHCAHTOA
anonymous
  • anonymous
???
mathmate
  • 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
  • mathmate
Hint: use the diagram I extracted from @Triciaal 's post.
anonymous
  • anonymous
hang on
anonymous
  • anonymous
I really have no idea... ummm did I miss cos?
mathmate
  • mathmate
more hint: |dw:1432904684712:dw| Pythagoras theorem says \(a^2+b^2=c^2\)
anonymous
  • anonymous
yes I know that
mathmate
  • 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
  • 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
  • mathmate
But now you understand why this is true?
anonymous
  • anonymous
yeah more or less
anonymous
  • anonymous
im leaning toward less
anonymous
  • anonymous
okay...so I know we have to use a reciprocal identity
mathmate
  • 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
  • anonymous
okay....
mathmate
  • 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
  • mathmate
If we can show that the left side is identical to the right, we're done.
anonymous
  • 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
  • mathmate
Exactly, that's an excellent start. A lot of the time, this strategy works!!!
anonymous
  • anonymous
yay!! okay so csc(theta) - 1/csc(theta) = cot(theta) cos(theta)
mathmate
  • mathmate
I don't see your reasoning, can you explain?
anonymous
  • anonymous
I replaced sin...so I was just filling that in the equation
mathmate
  • 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
  • mathmate
Are you at ease with adding fractions, such as 1/3 + 3?
anonymous
  • anonymous
oh! I put it on the wrong side yeah those are fine
mathmate
  • mathmate
1/3-3 = 1/3 - 3^2/3 = (1-3^2)/3 Does this make sense to you?
anonymous
  • anonymous
okay.... oh yes! I see what you did okay
mathmate
  • 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
  • anonymous
umm okay...
mathmate
  • mathmate
R U stuck or R U working on it?
anonymous
  • anonymous
trying to work on it...but im still stuck
mathmate
  • 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
  • anonymous
okay....so that's how you verify it?
mathmate
  • 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
  • mathmate
With what you know in terms of trigonometric identities, you should be able to finish the job, can you not?
anonymous
  • anonymous
csc(x) - sin(x) = 1?
mathmate
  • mathmate
Anyway, from the identity \(sin^2(x)+cos^2(x)=1\) we conclude: \(1-sin^2(x)=cos^2(x)\)
mathmate
  • mathmate
Does that help?
anonymous
  • anonymous
ohhhh...oops sorry... OHHH!!! I get it! yes that helps a lot!
anonymous
  • anonymous
Thank you so much for your help/time!!
mathmate
  • 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
  • anonymous
okay thank you. I think I can do that... I have to go, but thank you again!
mathmate
  • mathmate
You're welcome! :)
triciaal
  • triciaal
sorry system was down @mathmate thank you

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