Where Can I Find Help with Trig Differentiation, Limits, and Identities?

[Mspike6]

Hey guys..
first, i want to ask if anyone know a good tutorial or website that can help me with the Treg. deffrentiation, and limits...cause am troubles with that :(
I got some questions..

1)

Prove the Identity Tanx+1/Tanx-1 = Secx-cscx

I have no clue how to prove sucha a thing..



2)Evaluate Lim_{x-> Pi/4}1-Tanx/Sinx-Cosx


I tried to find any identity to change it's form, but i couldn;t figure anyone out


And the last one is Deffrentiate Sin2x=2sinxcosx, to devolop the identity of Cos2x


Y= Sin2x=2sinxcosx
y'=Cos2x.2 = -2cosxsinx
y'= Cos2x=-sinxcosx

But Cos2x should be equal to Cos²x-sin²x ??

 

[Mark44]

Hey guys..
first, i want to ask if anyone know a good tutorial or website that can help me with the Treg. deffrentiation, and limits...cause am troubles with that :(
I got some questions..

1)

Prove the Identity Tanx+1/Tanx-1 = Secx-cscx

You probably need parentheses on the left side. Otherwise, you have (tanx) +(1/tanx) - 1, and that's probably not what you meant.

I have no clue how to prove sucha a thing..



2)Evaluate Lim_{x-> Pi/4}1-Tanx/Sinx-Cosx

Again, you probably need parentheses. Otherwise, you have (1) - (tanx/sinx) - cosx, and that's probably not what you meant.

I tried to find any identity to change it's form, but i couldn;t figure anyone out


And the last one is Deffrentiate Sin2x=2sinxcosx, to devolop the identity of Cos2x

That's differentiate. Also, the word is trigonometry, not tregonometry.

Y= Sin2x=2sinxcosx
y'=Cos2x.2 = -2cosxsinx

In the equation above, you have correctly differentiated sin(2x) but not 2sinxcosx. This expression is a product, so you need to use the product rule.

y'= Cos2x=-sinxcosx

But Cos2x should be equal to Cos²x-sin²x ??

No, but the derivatative of cos(2x) should be -2sinxcosx = -2sin(2x)

 

[Mspike6]

Am sorry about that Mark

1)

Prove the Identity \frac{Tanx+1}{Tanx - 1}=\frac{Secx+cscx}{Secx-cscx}

2)Evaluate Lim_{x-> Pi/4}\frac{1-Tanx}{Sinx-Cosx}

and thanks for answering number 3 :D

 

[Mark44]

For 1, I would change everything to be in terms of sine and/or cosine.
LHS~=~ \frac{\frac{sinx}{cosx} + 1}{\frac{sinx}{cosx} - 1}

RHS~=~\frac{\frac{1}{cosx} + \frac{1}{sinx}}{\frac{1}{cosx} - \frac{1}{sinx}}

Now multiply the left-hand sid (LHS) by 1 in the form of cosx/cosx to simplify it, and multiply the RHS by 1 in the form of (cosx*sinx)/(cosx*sinx). If you end up with LHS = RHS, you will have proved the identity.

For 2, rewrite the numerator as 1 - (sinx/cosx), and then multiply the whole expression by 1 in the form of cosx/cosx. You should be able to take the limit then.

 

[singular]

For problem 1, ignore the RHS (left-hand side) for now.

Only look at \frac{tanx+1}{tanx - 1}. What fundamental identities involve tangent?

tanx=\frac{sinx}{cosx} and tanx=\frac{1}{cotx}

Any others?

What about fundamental identities that involve secant (sec) and cosecant (csc)?

Use some of the fundamental identities and see if you can get the LHS to contain only secants and cosecants.

Once you are there, you will need a little trick. If you can't figure that part out, come back and ask for more help.

 

[Mspike6]

Thanks for helping me guys... i got number 1 right

For 2, rewrite the numerator as 1 - (sinx/cosx), and then multiply the whole expression by 1 in the form of cosx/cosx. You should be able to take the limit then.

after doing that i got \frac{Cosz-Sinx}{SinxCosx-Cos^2 x}

and when i take the limit, the dominator still goes to 0

 

[singular]

Thanks for helping me guys... i got number 1 right



after doing that i got \frac{Cosx-Sinx}{SinxCosx-Cos^2 x}

and when i take the limit, the dominator still goes to 0

You have a good start.

Try L'Hôpital's rule (use only in cases where you have 0/0 or ∞/∞)

lim_{n\rightarrow c}\frac{f(x)}{g(x)}=lim_{n\rightarrow c}\frac{f'(x)}{g'(x)}

Then evaluate it and see what you get.

 

[Mark44]

Thanks for helping me guys... i got number 1 right



after doing that i got \frac{Cosz-Sinx}{SinxCosx-Cos^2 x}

and when i take the limit, the dominator still goes to 0

Factor the denominator and you'll get some cancellation with the numerator.

 

[Mark44]

You have a good start.

Try L'Hôpital's rule (use only in cases where you have 0/0 or ∞/∞)

lim_{n\rightarrow c}\frac{f(x)}{g(x)}=lim_{n\rightarrow c}\frac{f'(x)}{g'(x)}

Then evaluate it and see what you get.

L'Hopital's Rule is unnecessary for this problem. Also, this is the Precalculus forum, so a calculus technique that requires the knowledge of derivatives is probably inappropriate. OTOH, the OP might have mistakenly posted here.

 

[singular]

L'Hopital's Rule is unnecessary for this problem. Also, this is the Precalculus forum, so a calculus technique that requires the knowledge of derivatives is probably inappropriate. OTOH, the OP might have mistakenly posted here.

I only assumed because of the use of limits, which in my experience are taught in calculus.

Mspike6: I was able to check the method Mark44 is trying to help you with, so if you want to see a way that takes only a little extra work with no calculus, but ends up much simpler in the end, try to follow his tips.

 

[Mspike6]

Sorry for taking me to long to replay.

the course am taking is the course right befor Calculas I , It's called Pre-Calculas .

I did it the way Mark said it worked perfectly.

Thank you both Mark anf Singular..

But i would like to ask again...do you, by any chance, know a tutorial or Website that can help me with this Kinda stuff ? Like Trig Limit, Trig Identeties and such .

Thank you

 

[singular]

tutorial or Website that can help me with this Kinda stuff ? Like Trig Limit, Trig Identeties and such

Here is a nice http://www.pinkmonkey.com/studyguides/subjects/calc/chap2/c0202701.asp" .

The http://en.wikipedia.org/wiki/List_of_trigonometric_identities" on trig identities is good, but it probably has more than you will ever need.

Here is a good condensed http://www.sosmath.com/trig/Trig5/trig5/trig5.html" .