arccos0?
1/cos(0)
\[\tan ^{-1}-\frac{ \sqrt{3} }{ 3}\]
@tomo is that the asnwer?
?? that doesn't make sense. please reword
the one with tangent?
what is arccos0??
cos(0) = 1.
@tomo
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?
@Inspired
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?
@Inspired
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|
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