arccos0?

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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1/cos(0)
\[\tan ^{-1}-\frac{ \sqrt{3} }{ 3}\]
@tomo is that the asnwer?

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Other answers:

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

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