Question

how do i verify

cot (x- pi/2)=-tanx

i dont know how to do this so pls be patient with me!

Answer 1

hmmm \(\bf cot\left( x\cfrac{\pi }{2} \right) = -tan(x)?\)

Answer 2

cot (x- pi/2)=-tanx

Answer 3

sorry about that!

Answer 4

k

Answer 5

hmmm what's the \(\bf cos\left(\frac{\pi }{2} \right)\quad and \quad sin\left(\frac{\pi }{2} \right)\)

Answer 6

what?

Answer 7

ehehh, what are, check your Unit Circle, the \(\bf cos\left(\frac{\pi }{2} \right)\quad and \quad sin\left(\frac{\pi }{2} \right)?\)

Answer 8

-1/2 and 0 ?

Answer 9

-1/2 and 0? check your unit circle closer

Answer 10

(0,1) ? i dont know i just started

Answer 11

hmmm, well, \(\bf cos\left(\frac{\pi }{2} \right)=0\quad and \quad sin\left(\frac{\pi }{2} \right)=1\)

so we know that much

Answer 12

how is that relevant?

Answer 13

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

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

Answer 14

wait wait wait how did you get that huge second fraction

Answer 15

recall \(\bf \textit{Sum and Difference Identities}


\\ \quad \\

sin({\color{brown}{ \alpha}} - {\color{blue}{ \beta}})=sin({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}})- cos({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})




\\ \quad \\

cos({\color{brown}{ \alpha}} - {\color{blue}{ \beta}})= cos({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})\)

Answer 16

a=x and b=pi/2 ???

Answer 17

yeap

Answer 18

so sin/-cos=-tan because tan=sin/cos right

Answer 19

\(\bf \textit{Sum and Difference Identities}


\\ \quad \\

sin({\color{brown}{ x}} - {\color{blue}{ \frac{\pi }{2}}})=sin({\color{brown}{ x}})cos({\color{blue}{ \frac{\pi }{2}}})- cos({\color{brown}{ x}})sin({\color{blue}{ \frac{\pi }{2}}})




\\ \quad \\

cos({\color{brown}{ x}} - {\color{blue}{ \frac{\pi }{2}}})= cos({\color{brown}{ x}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ x}})sin({\color{blue}{ \frac{\pi }{2}}})\)

Answer 20

yeap

Answer 21

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

Answer 22

thank you sooo much

Answer 23

yw