A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
how do i verify
cot (x pi/2)=tanx
i dont know how to do this so pls be patient with me!
anonymous
 one year ago
how do i verify cot (x pi/2)=tanx i dont know how to do this so pls be patient with me!

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm \(\bf cot\left( x\cfrac{\pi }{2} \right) = tan(x)?\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm what's the \(\bf cos\left(\frac{\pi }{2} \right)\quad and \quad sin\left(\frac{\pi }{2} \right)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ehehh, what are, check your Unit Circle, the \(\bf cos\left(\frac{\pi }{2} \right)\quad and \quad sin\left(\frac{\pi }{2} \right)?\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01/2 and 0? check your unit circle closer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(0,1) ? i dont know i just started

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm, well, \(\bf cos\left(\frac{\pi }{2} \right)=0\quad and \quad sin\left(\frac{\pi }{2} \right)=1\) so we know that much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how is that relevant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf cot\left( x\frac{\pi }{2} \right)\implies \cfrac{cos\left( x\frac{\pi }{2} \right)}{sin\left( x\frac{\pi }{2} \right)} \\ \quad \\ \cfrac{cos(x)cos\left(\frac{\pi }{2} \right)+sin(x)sin\left(\frac{\pi }{2} \right)}{sin(x)cos\left(\frac{\pi }{2} \right)cos(x)sin\left(\frac{\pi }{2} \right)} \\ \quad \\ \cfrac{cos(x)\cdot {\color{brown}{ 0}}+sin(x)\cdot {\color{brown}{ 1}}}{sin(x)\cdot {\color{brown}{ 0}}cos(x)\cdot {\color{brown}{ 1}}}\implies \cfrac{sin(x)}{cos(x)}\implies ?\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait wait wait how did you get that huge second fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0recall \(\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({\color{brown}{ \alpha}}  {\color{blue}{ \beta}})=sin({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}}) cos({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}}) \\ \quad \\ cos({\color{brown}{ \alpha}}  {\color{blue}{ \beta}})= cos({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so sin/cos=tan because tan=sin/cos right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({\color{brown}{ x}}  {\color{blue}{ \frac{\pi }{2}}})=sin({\color{brown}{ x}})cos({\color{blue}{ \frac{\pi }{2}}}) cos({\color{brown}{ x}})sin({\color{blue}{ \frac{\pi }{2}}}) \\ \quad \\ cos({\color{brown}{ x}}  {\color{blue}{ \frac{\pi }{2}}})= cos({\color{brown}{ x}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ x}})sin({\color{blue}{ \frac{\pi }{2}}})\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \cfrac{cos(x)\cdot {\color{brown}{ 0}}+sin(x)\cdot {\color{brown}{ 1}}}{sin(x)\cdot {\color{brown}{ 0}}cos(x)\cdot {\color{brown}{ 1}}}\implies \cfrac{sin(x)}{cos(x)}\implies \cfrac{sin(x)}{1\cdot cos(x)} \\ \quad \\ \cfrac{1}{1}\cdot \cfrac{sin(x)}{cos(x)}\implies  \cfrac{sin(x)}{cos(x)}\implies tan(x)\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.