Question

Verify the identity.
cot(x-pi/2) = -tanx

Answer 1

Ok?

Answer 2

Sorry it didn't let me type it all out. cot(x-pi/2) = -tanx

Answer 3

blank, why? because "blank" = "blank"

Answer 4

how do I go about proving it?

Answer 5

\(\bf {\color{blue}{ cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}}}\\ \quad \\ \quad \\
cot\left(x-\frac{\pi}{2}\right)=-tan(x)\\ \quad \\

cot\left(x-\frac{\pi}{2}\right)\implies \cfrac{cos\left(x-\frac{\pi}{2}\right)}{sin\left(x-\frac{\pi}{2}\right)}\\ \quad \\
\implies \cfrac{cos(x)cos\left(\frac{\pi}{2}\right)+sin(x)sin\left(\frac{\pi}{2}\right)}{sin(x)cos\left(\frac{\pi}{2}\right)-cos(x)sin\left(\frac{\pi}{2}\right)}\)

notice your Unit Circle, what's \(\bf cos\left(\frac{\pi}{2}\right)\quad ?\)
what about the \(\bf sin\left(\frac{\pi}{2}\right)\quad ?\)

Answer 6

Isn't cos 1/2?

Answer 7

what do you mean?

Answer 8

the cos (pi/2). is it 1/2?

Answer 9

is that what you get from the Unit Circle?

Answer 10

I'm trying to recall a chart we made in class regarding this. And I believe that it is 1/2

Answer 11

well... you may want to get a Unit Circle :) if you don't have one
there are many online btw

Answer 12

on the unit circle it says pi/2 is 90 degrees and it is located at (0,1)

Answer 13

ok.... so.... the cosine... value will be?

Answer 14

1

Answer 15

http://www.sciencedigest.org/UnitCircle.gif <---- check this unit circle, it has the pair as cosine,sine

keep in mind that cosine is just the ADJACENT SIDE's length
and sine is the OPPOSITE SIDE's length on the given angle

(x,y) => (cosine, sine)

Answer 16

Oh alright I had it backwards. So cos is 0 and sin is 1. What would be the next step in proving the identity?

Answer 17

ok... so now we know that \(\bf cos\left(\frac{\pi}{2}\right)=0\ and\ sin\left(\frac{\pi}{2}\right)=1\)


thus \(\bf \cfrac{cos(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}+sin(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}
{sin(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}-cos(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}\\ \quad \\

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

what would you get for that?

Answer 18

Are we trying to find the tangent? So x/y? or 0/1? Which would just be 0, correct?

Answer 19

verify the identity..... as in
make the left-side like the right-side or the other way around

Answer 20

\(\bf \cfrac{cos(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}+sin(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}
{sin(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}-cos(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}\\ \quad \\

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

Answer 21

looks closely

Answer 22

look rather... anyhow... so...what do you think?

Answer 23

I apologize, you're doing a very good job of explaining this, I'm just still completely lost.

Answer 24

recall that SOMETHING multiplied by 0 is = ?
and SOMETHING multiplied by 1 = ?

Answer 25

so it's 0?

Answer 26

lemme.... rewrite it some

Answer 27

\(\bf \cfrac{cos(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}+sin(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}
{sin(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}-cos(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}\\ \quad \\

\implies \cfrac{cos(x)\cdot {\color{blue}{0}}+sin(x)\cdot {\color{blue}{ 1}}}
{sin(x)\cdot {\color{blue}{ 0}}-cos(x)\cdot {\color{blue}{ 1}}}\qquad cos(x)=a\qquad sin(x)=b\\ \quad \\
\implies \cfrac{(a\cdot 0)-(b\cdot 1)}{(b\cdot 0)-(a\cdot 1)}\)

what would you get from that fraction?

Answer 28

1?

Answer 29

well.... how did you get 1?

Answer 30

hmmm actually it's supposed to be a + above... so \(\bf \cfrac{(a\cdot 0)+(b\cdot 1)}{(b\cdot 0)-(a\cdot 1)}\)

Answer 31

0 multiplied by anything is 0. So is the answer -1?

Answer 32

-1? is it? whatever happend to the variables?

Answer 33

hehehe, the variables went MIA =)

Answer 34

\[-\frac{ \sin }{ \cos }\]

Answer 35

ahhhh... notice.... \(\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad thus\qquad -\cfrac{sin(\theta)}{cos(\theta)}\implies -tan(\theta)\)

Answer 36

\(\bf \cfrac{cos(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}+sin(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}
{sin(x){\color{blue}{ cos\left(\frac{\pi}{2}\right)}}-cos(x){\color{blue}{ sin\left(\frac{\pi}{2}\right)}}}\\ \quad \\


\implies \cfrac{cos(x)\cdot {\color{blue}{0}}+sin(x)\cdot {\color{blue}{ 1}}}
{sin(x)\cdot {\color{blue}{ 0}}-cos(x)\cdot {\color{blue}{ 1}}}\implies -\cfrac{sin(x)}{cos(x)}\implies -tan(x)\)

Answer 37

Wow okay! So that is how the proof is from start to finish?

Answer 38

Thank you so much! You're awesome!