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

- anonymous

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

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- freckles

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

- freckles

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

- anonymous

can u guide me @freckles

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## More answers

- freckles

which part do you need help on?

- freckles

did you draw the right triangle yet?

- anonymous

i guess all.

- anonymous

yes

- freckles

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

- anonymous

ok

- freckles

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

- anonymous

no

- freckles

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

- anonymous

adj is 1 and hyp is 2?

- freckles

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

- anonymous

ok

- freckles

have you solved for opp yet?

- freckles

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

- anonymous

still working on it

- anonymous

opp is root 5?

- freckles

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

- freckles

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

- anonymous

shoot my bad

- freckles

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

- freckles

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

- anonymous

ok

- anonymous

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

- anonymous

@freckles

- freckles

what happen to the sqrt( ) part ?

- freckles

also you should have two answers for each

- freckles

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

- anonymous

ok

- anonymous

±root 3?

- freckles

yep sounds great

- anonymous

how did you get the root sighn tho

- anonymous

sign

- anonymous

jk never mind

- freckles

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

- freckles

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

- anonymous

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

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