A community for students.
Here's the question you clicked on:
 0 viewing
klbala2006
 3 years ago
sin(sin^{1} \frac{ sqrt{3} }{ 2 }+cos^{1} \frac{ 1 }{ 2 })
klbala2006
 3 years ago
sin(sin^{1} \frac{ sqrt{3} }{ 2 }+cos^{1} \frac{ 1 }{ 2 })

This Question is Open

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0@klbala2006 Do you know the range of \(\sin^{1} x\) ?

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.0looking at it its all 4th quadrant dw:1352871092096:dw

klbala2006
 3 years ago
Best ResponseYou've already chosen the best response.0i need to evaluate this function . @ash2326 no i dont

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know the range and domain of sin (x) ?

klbala2006
 3 years ago
Best ResponseYou've already chosen the best response.0@ash2326 domain of sin(x) is (R), while its range is [1,1]

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0Good, to have the inverse of sin function, we limit it's range domain= range of sin x= [1, 1] Range = \(\large [\frac{\pi}{2}, \frac{\pi}{2}]\) DO you get this?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0if x is positive, then y lies in \([0, \frac {\pi}{2}]\) if x is negative, then y lies in \([\frac{\pi}{2}, 0)\) Now what's \(\sin^{1}(\frac{\sqrt 3}{2})\) ?

klbala2006
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \pi }{ 4 }\] in radian

klbala2006
 3 years ago
Best ResponseYou've already chosen the best response.0@ash2326 complete please

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, I was away. But sin (45) is not \(\frac{\sqrt 3}{2}\) it's, \(\frac{1}{\sqrt 2}\).
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.