Question

helppp

Answer 1

4.

arccos (-1/2) = x

cos x = -1/2

x=2pi/3

Answer 2

arcsin(sqrt(2)/2) = x

sin x = sqrt(2)/2

What do you think x is?

Answer 3

i thought the answer was pi but i got it wrong , this is really frustrating

Answer 4

Think of the sine graph. Where is sin x = sqrt(2)/2 ?

the range of the arcsin function is between [-pi/2,pi/2] so the answer lies within that region

Answer 5

i honestly dont know, im sorry my teacher is really bad at explaining. I am trying to use the book as a guide but i find it really complicated without anyone's help

Answer 6

Hint:

\[{\sqrt2 \over 2 } = {1 \over \sqrt 2}\]

sin x = 1/sqrt2

Does that look familiar?

Answer 7

ok so the answer is pi/4?

Answer 8

yep :)

Answer 9

got it! thank youu

Answer 10

for arccos0 the answer is pi/2 right?

Answer 11

yeah :) nice!

Answer 12

Firstly, they give you a key point: [-3,pi]

This is important, since we know the domain of the arccos and arcsin functions are between -1 and 1.

arctan looks a lot different than that graph, so you can rule that out.

so it must be either arcsin(t + 2) or arccos(t+2)

We know that arcsins range is [-pi/2,pi/2]

and arccos range is [0,pi]

which leaves us with 2. arccos(t+2)

Answer 13

it would be 1?

Answer 14

yeah ! You are owning this!

Answer 15

haha i have one that i really dont understand

Answer 16

which one do you struggle with?

Answer 17

a, b or c?

Answer 18

all of them actually

Answer 19

well not a

Answer 20

do you know the Pythagoras theorem?

Answer 21

yeah i use that for a right?

Answer 22

ah cool. what did you get for a?

Answer 23

21

Answer 24

Cool, that's correct. Now for b and c:

You have to use the cosine law for b.

Do you know the formula?

Answer 25

http://en.wikipedia.org/wiki/Law_of_cosines

Answer 26

i did sin of A so 15/26 and i got a decimal

Answer 27

What angle did you use though?