ATAR Notes: Forum

Uni Stuff => Universities - Victoria => University of Melbourne => Topic started by: VivaTequila on June 02, 2012, 04:18:37 pm

Title: Calc 1 Trig Substitution
Post by: VivaTequila on June 02, 2012, 04:18:37 pm: Asked to simplify: Cos^2(x)/sin(x) + sin(x)
Answer is cosec^2(x)

I understand that if you divide both terms by sin(x), you get:
cot(x)+1

which is equivalent to cosec^2(x)

but why is it that when you multiply both terms by sin(x), you get cos^2(x)+sin^2(x), which is 1, which is not cosec^2(x) - the answer...

cheers
Title: Re: Calc 1 Trig Substitution
Post by: b^3 on June 02, 2012, 04:27:07 pm: Are you sure $\csc^{2}(x)$ is the answer?
Wouldn't it just be $\csc(x)$ ?
$\begin{alignedat}{1}\frac{\cos^{2}(x)}{\sin(x)}+\sin(x) & =\frac{\cos^{2}(x)+\sin^{2}(x)}{\sin(x)}<br />\\ & =\frac{1}{\sin(x)}<br />\\ & =\csc(x)<br />\end{alignedat}$

Quote from: VivaTequila on June 02, 2012, 04:18:37 pm

I understand that if you divide both terms by sin(x), you get:
cot(x)+1

which is equivalent to cosec^2(x)
Also note that when you divide by sin(x), that $\cot(x)+1\neq\csc^{2}(x)$ , it's $\cot^{2}(x)+1=\csc^{2}(x)$

Also just to make sure I'm not seeing things, http://www.wolframalpha.com/input/?i=cos%5E2(x)%2Fsin(x)%2Bsin(x)&t=crmtb01

So the answer is $\csc(x)$ not $\csc^{2}(x)$
Title: Re: Calc 1 Trig Substitution
Post by: VivaTequila on June 02, 2012, 05:00:09 pm: right you are, thanks for the speedy reply!

and hmm...

why is it that the sin(x) doesn't cancel out?

ATAR Notes | SMF © 2014, Simple Machines