anonymous
  • anonymous
arccos0?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1/cos(0)
anonymous
  • anonymous
\[\tan ^{-1}-\frac{ \sqrt{3} }{ 3}\]
anonymous
  • anonymous
@tomo is that the asnwer?

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anonymous
  • anonymous
?? that doesn't make sense. please reword
anonymous
  • anonymous
the one with tangent?
anonymous
  • anonymous
what is arccos0??
anonymous
  • anonymous
cos(0) = 1.
anonymous
  • anonymous
@tomo
anonymous
  • anonymous
arccos(0)= 90 degrees or pi/2
anonymous
  • anonymous
one second. it is a little different than i first thought it was.
anonymous
  • anonymous
k but plz try to expalin i have test on this toorrow
anonymous
  • anonymous
k what is the answer of the questio with tangent?
anonymous
  • anonymous
@Inspired
anonymous
  • 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
anonymous
  • anonymous
cos(90) is 0*
anonymous
  • 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)
anonymous
  • 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
anonymous
  • anonymous
for the test tomorrow
anonymous
  • anonymous
socan u tell me the answer to the tangent one?plz
anonymous
  • anonymous
\[\arcsin-\frac{ \sqrt{3} }{ 2 }\]
anonymous
  • anonymous
and this one plz i have to go emergency plz answe those 2 for me plz
anonymous
  • anonymous
that will be really helpful
anonymous
  • anonymous
-pi/6 and -pi/3 (take a look at the unit circle for the second one)
anonymous
  • 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?
anonymous
  • anonymous
@Inspired
anonymous
  • 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|
anonymous
  • 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.
anonymous
  • anonymous
but is is y/x right why it doesn't work
anonymous
  • anonymous
if u do it with the values on unit square and i thought the domain is from (pi/2,-pi/2)?
anonymous
  • anonymous
|dw:1358819636475:dw|
anonymous
  • 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
anonymous
  • anonymous
:)
anonymous
  • anonymous
\[COS ^{-1}(\sin 11\pi/6)\]
anonymous
  • anonymous
@Inspired
anonymous
  • anonymous
@Inspired last one
anonymous
  • anonymous
and which one do u do first?
anonymous
  • 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)
anonymous
  • anonymous
is that the answer arccos(-1/20)

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