Question

Verify the identity. Cot[theta-(pi/2)=-tan(theta)

Answer 1

\(\bf cos\left(\theta-\frac{\pi}{2}\right)=-tan(\theta)
\\ \quad \\ \quad \\
cos\left(\theta-\frac{\pi}{2}\right)\implies cos(\theta)cos\left(\frac{\pi}{2}\right)+sin(\theta)sin\left(\frac{\pi}{2}\right)
\\ \quad \\
\textit{what is the }sin\left(\frac{\pi}{2}\right)\quad ?
\\ \quad \\
\textit{what is the }cos\left(\frac{\pi}{2}\right)\quad ?\)

Answer 2

hmmm well. it helps if I used cotangent identity, rather than cosine =)

hehe lemme rewrite it some

Answer 3

\(\bf cot\left(\theta-\frac{\pi}{2}\right)=-tan(\theta)
\\ \quad \\ \quad \\
cot\left(\theta-\frac{\pi}{2}\right)\implies \cfrac{1}{tan\left(\theta-\frac{\pi}{2}\right)}\implies
\cfrac{1}{\frac{tan(\theta)-tan\left(\frac{\pi}{2}\right)}{1+tan(\theta)tan\left(\frac{\pi}{2}\right)}}
\implies
\cfrac{1}
{
\frac{\frac{sin(\theta)}{cos(\theta)}-\frac{sin\left(\frac{\pi}{2}\right)}{cos\left(\frac{\pi}{2}\right)}}{1+\frac{sin(\theta)}{cos(\theta)}\frac{sin\left(\frac{\pi}{2}\right)}{cos\left(\frac{\pi}{2}\right)}}
}
\\ \quad \\
\textit{what is the }sin\left(\frac{\pi}{2}\right)\quad ?
\\ \quad \\
\textit{what is the }cos\left(\frac{\pi}{2}\right)\quad ?\)

Answer 4

Uuuuuuhhhhhhh.......

Answer 5

lemme recheck myself

Answer 6

It might not be you, I'm just not good at this algebra/trig stuff.

Answer 7

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/trid1.gif <--- is all we're using, the tangent identity

Answer 8

Okay, I figured that much, but I need help doing most of the steps in general. I just have no idea where to go from here.

Answer 9

hmmm actually... that didn't yield much ....

Answer 10

lemme check something

Answer 11

gimme a few secs

Answer 12

Okay

Answer 13

\(\bf cot\left(\theta-\frac{\pi}{2}\right)=-tan(\theta)
\\ \quad \\ \quad \\

cot\left(\theta-\frac{\pi}{2}\right)\implies \cfrac{cos\left(\theta-\frac{\pi}{2}\right)}{sin\left(\theta-\frac{\pi}{2}\right)}
\implies
\cfrac{cos(\theta)cos\left(\frac{\pi}{2}\right)+sin(\theta)sin\left(\frac{\pi}{2}\right)}{sin(\theta)cos\left(\frac{\pi}{2}\right)-cos(\theta)sin\left(\frac{\pi}{2}\right)}
\\ \quad \\
\textit{what is the }sin\left(\frac{\pi}{2}\right)\quad ?
\\ \quad \\
\textit{what is the }cos\left(\frac{\pi}{2}\right)\quad ?\)

Answer 14

you should be able to get those from the Unit Circle

Answer 15

Okay, uuummm. Still not sure. I did it on my calculator and it said sin(pi/2)=1 and cos(pi/2)=0. I'm at a double disadvantage because I'm naturally bad at math, and I don't think these concepts were explained well in my text book.

Answer 16

ahemm well
do you happen to have a Unit Circle? many online if you need one

Answer 17

Yeah, there's a unit circle.

Answer 18

but yes, the calculator is correct
check your Unit Circle


so we know that \(\bf sin\left(\frac{\pi}{2}\right)\implies 1\qquad cos\left(\frac{\pi}{2}\right)\implies 0\qquad thus
\\ \quad \\
\cfrac{cos(\theta){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}+sin(\theta){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}{sin(\theta){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}-cos(\theta){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}\implies

\cfrac{cos(\theta){\color{blue}{0}}+sin(\theta){\color{blue}{1}}}{sin(\theta){\color{blue}{ 0}}-cos(\theta){\color{blue}{1}}}\)

so what does that leave you with?

Answer 19

0+1=1. 0-1=-1. 1/-1=-1 ?

Answer 20

? whatever happened to \(\bf sin(\theta)\ and \ cos(\theta)\quad ?\)

Answer 21

Uuuuuhhhhh, they went on holiday? Um I really don't know. I get the sense that I should know this, but I'm kinda clueless. Sorry for my ignorance. I'm looking at a unit circle online, I'm hoping it will be more informative than my text book.

Answer 22

hehe... lemme rewrite that

Answer 23

thought surely they'd be happy to go on holiday =)

Answer 24

\(\bf sin\left(\frac{\pi}{2}\right)\implies 1\qquad cos\left(\frac{\pi}{2}\right)\implies 0\qquad thus
\\ \quad \\
\cfrac{cos(\theta)\cdot {\color{blue}{ cos\left(\frac{\pi}{2}\right)}}+sin(\theta)\cdot {\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}{sin(\theta)\cdot {\color{blue}{ cos\left(\frac{\pi}{2}\right)}}-cos(\theta)\cdot {\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}\implies

\cfrac{cos(\theta)\cdot {\color{blue}{0}}+sin(\theta)\cdot {\color{blue}{1}}}{sin(\theta)\cdot {\color{blue}{ 0}}-cos(\theta)\cdot {\color{blue}{1}}}\)


so... what would you be left with?

Answer 25

My jacked-up-math-rejecting-brain is telling me that answer is -[sin(theta)/cos(theta)]

Answer 26

yeap
\((\bf \cfrac{cos(\theta)\cdot {\color{blue}{0}}+sin(\theta)\cdot {\color{blue}{1}}}{sin(\theta)\cdot {\color{blue}{ 0}}-cos(\theta)\cdot {\color{blue}{1}}}\implies \cfrac{sin(\theta)}{-cos(\theta)}\implies -\cfrac{sin(\theta)}{cos(\theta)}\)

does that look like ... .an identity ?

Answer 27

\(\bf -\cfrac{sin(\theta)}{cos(\theta)}\implies -1\cdot {\color{blue}{ \cfrac{sin(\theta)}{cos(\theta)}}}\)

Answer 28

Okay, so is that what I'm looking for? -1 * sine(theta)/cos(theta)?

Answer 29

http://www.efunda.com/math/trig_functions/images/trig_relation.gif <--- any of those look like it ?

Answer 30

Yeah, first one at the top. tan=sinx/cosx

Answer 31

ahhh well, there

Answer 32

\(\bf \bf -\cfrac{sin(\theta)}{cos(\theta)}\implies -1\cdot {\color{blue}{ \cfrac{sin(\theta)}{cos(\theta)}}}\implies -{\color{blue}{ tan(\theta)}}
\)

Answer 33

Oh, cool. I feel like I've been tricked into getting the correct answer, but I'll take it. Thanks. But, stay available for a few minutes, will ya, I've got try and name the steps that get me from one step to the next, they're kinda strict about showing your work.

Answer 34

well... sure...
you can just check above, is pretty straightforward

Answer 35

Cool, thanks.