## anonymous 3 years ago arccos0?

1. anonymous

1/cos(0)

2. anonymous

$\tan ^{-1}-\frac{ \sqrt{3} }{ 3}$

3. anonymous

@tomo is that the asnwer?

4. anonymous

?? that doesn't make sense. please reword

5. anonymous

the one with tangent?

6. anonymous

what is arccos0??

7. anonymous

cos(0) = 1.

8. anonymous

@tomo

9. anonymous

arccos(0)= 90 degrees or pi/2

10. anonymous

one second. it is a little different than i first thought it was.

11. anonymous

k but plz try to expalin i have test on this toorrow

12. anonymous

k what is the answer of the questio with tangent?

13. anonymous

@Inspired

14. anonymous

Well it's just the inverse of cos(0). Cos(0) =1. The inverse of that switches the values of the sine cosine functions. cos(90) gives you 1. but because remember, sin(0)=0 and sin(90)=1

15. anonymous

cos(90) is 0*

16. anonymous

and an easy way to remember is: sin is your y axis and cos is your x axis. the inverse of a function is basically switches the x and y. what was y=f(x) is now x=f(y)

17. anonymous

k i know that but i have like 5 problems to work on i just want the right answers to it so i can practice

18. anonymous

for the test tomorrow

19. anonymous

socan u tell me the answer to the tangent one?plz

20. anonymous

$\arcsin-\frac{ \sqrt{3} }{ 2 }$

21. anonymous

and this one plz i have to go emergency plz answe those 2 for me plz

22. anonymous

23. anonymous

-pi/6 and -pi/3 (take a look at the unit circle for the second one)

24. anonymous

how did u get -pi/6? i know in the unit cirlce but when i tried all of them tan is y/x then which onw will work?

25. anonymous

@Inspired

26. anonymous

Let's see. arctan (sqrt3/3) is the same thing as arctan (1/sqrt3) right? and using 30-60-90 triangle, you have this: |dw:1358819026253:dw|

27. anonymous

Anyway, since the value is negative, it will occur in the 2nd or 4th Quadrant. Since we want a single value, you usually use the 4th quadrant, and that gives you a negative value.

28. anonymous

but is is y/x right why it doesn't work

29. anonymous

if u do it with the values on unit square and i thought the domain is from (pi/2,-pi/2)?

30. anonymous

|dw:1358819636475:dw|

31. anonymous

OMG K GET IT THANKS THANKS ALOT @Inspired U ARE AWESOME AND HELPFUL OMG THANKS THANKS I JUST FEEL SO HAPPY WHEN I GET IT THANKS

32. anonymous

:)

33. anonymous

$COS ^{-1}(\sin 11\pi/6)$

34. anonymous

@Inspired

35. anonymous

@Inspired last one

36. anonymous

and which one do u do first?

37. anonymous

ok :0 sin(11pi/6) is the same thing as sin(12pi/6-pi/6) so you get sin(2pi-pi/6) --> -sin(pi6)=-1/2 and so you have arccos(-1/2)

38. anonymous