Question

Find the Value.

Answer 1

\[\sin(\cos ^{-1}(\frac{ \sqrt2 }{ 2 }))\]

Answer 2

you working in degrees or radians?

Answer 3

may i have the solution in radians please

Answer 4

well, ide actually like more than jsut the solution, how to do it is more important :p

Answer 5

the answer will not be in degrees or radians, the answer will be a number

Answer 6

ok

Answer 7

them im guessing i need to convert it to radians first

Answer 8

but i asked because working from the inside out the first question is what is
\[\cos^{-1}(\frac{\sqrt2}{2})\] i.e. what angle has cosine \(\frac{\sqrt2}{2}\)

Answer 9

pi/4

Answer 10

if you are working in degrees, the answer is 45
if you are working in radians, the answer is \(\frac{\pi}{4}\) ok good we got that

Answer 11

so all that is left is to compute
\[\sin(\frac{\pi}{4})\]

Answer 12

wait
cos(sqrt2/2) = pi/4

Answer 13

oh no lets go back slowly

Answer 14

what about arccos sqt2/2

Answer 15

don't get confused here
\[\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}\] right?

Answer 16

yes

Answer 17

that makes
\[\cos^{-1}(\frac{\sqrt2}{2})=\frac{\pi}{4}\]

Answer 18

so
\[\sin(\cos^{-1}(\frac{\sqrt2}{2}))=\sin(\frac{\pi}{4})\]

Answer 19

so its the same thing? thats whats confusing me

Answer 20

and
\[\sin(\frac{\pi}{4})=\frac{\sqrt2}{2}\] as your final answer

Answer 21

shouldn't be too confusing since there is a well known point on the unit circle where both sine and cosine are both \(\frac{\sqrt2}{2}\)

Answer 22

ahhhh ok i think i got it

Answer 23

hope it is more or less clear

Answer 24

So,
\[\arccos (\frac{ \sqrt2 }{ 2 }) = ?\]
\[\cos (?) = \frac{ \sqrt 2 }{ 2 }\]

Answer 25

zactly

Answer 26

thankkkk youuuu

Answer 27

so easy it blows my mind

Answer 28

yw
now i have a question

Answer 29

shoot

Answer 30

how come you got that stern grumpy face even though you go to school in hawaii?

Answer 31

hahaha because math is hard!

Answer 32

ok that's fair
good luck with it

Answer 33

thank you so much

Answer 34

yw