Question

If cosine of x equals 1 over 2, what is sin(x) and tan(x)? Explain your steps in complete sentences.

Answer 1

make a right triangle
then use cos(x)=1/2
then use Pythagorean theorem to figure out opposite side of x

Answer 2

then you have everything you need to find sin(x) and tan(x)

Answer 3

can u guide me @freckles

Answer 4

which part do you need help on?

Answer 5

did you draw the right triangle yet?

Answer 6

i guess all.

Answer 7

yes

Answer 8

ok then use cos(x)=1/2
remember cosine, think adjacent/hyp

Answer 9

ok

Answer 10

then use the Pythagorean theorem to find the opp side
can you do that?

Answer 11

no

Answer 12

well we know the Pythagorean theorem says:
\[opp^2+adj^2=hyp^2 \\ \text{ and we know the } adj \text{ and the hyp } \\ \text{ just solve for opp}\]

Answer 13

adj is 1 and hyp is 2?

Answer 14

yes this was given by cos(x)=1/2
you could also use any thing that reduces down to 1/2
for example cos(x)=10/20
and call the adj=10 and hyp=20

but it doesn't matter in the end we will get the same reduced fractions

so yes you can use adj=1 and hyp=2

Answer 15

ok

Answer 16

have you solved for opp yet?

Answer 17

if you have you can find sin(x) and tan(x)

Answer 18

still working on it

Answer 19

opp is root 5?

Answer 20

\[oop^2+1^2=2^2 \\ oop^2+1=4\]
4-1 isn't 5
should be 3

Answer 21

\[opp^2=4-1 \\ opp^2=3 \\ opp= \pm \sqrt{3}\]

Answer 22

shoot my bad

Answer 23

so you have two answers for both sin(x) and tan(x)
because of the plus or minus thing

Answer 24

anyways recall sin(x)=opp/hyp
and tan(x)=opp/adj

Answer 25

ok

Answer 26

sin(x) = 3/2
tan(x) = 3/1 = 3

Answer 27

@freckles

Answer 28

what happen to the sqrt( ) part ?

Answer 29

also you should have two answers for each

Answer 30

\[\sin(x)=\frac{ \pm \sqrt{3}}{2} \text{ this is the two answers for } \sin(x) \]
now you try tan(x)=?

Answer 31

ok

Answer 32

±root 3?

Answer 33

yep sounds great

Answer 34

how did you get the root sighn tho

Answer 35

sign

Answer 36

jk never mind

Answer 37

\[opp^2+adj^2=hyp^2 \\ opp^2+1^2=2^2 \\ opp^2=4-1 \\ opp^2=3 \\ \text{ \to solve an equation like this you take square root of both sides }\]

Answer 38

but it will also mean you wind up with two possibilities
since sqrt(opp^2)=|opp|=opp if opp>0
but =-opp if opp<0

Answer 39

ok. Thank You so so so so much!!!!!!!!!!!!!!!!!!!