Question

--Please help--

Answer 1

If 2cos theta = root 3, find the value of 2 cos theta +1 / 2 cos theta - 1

Answer 2

\[2 \cos \theta + 1/ 2 \cos \theta - 1 = ?\]

Answer 3

\(2cos\theta=\sqrt{3}\)

divide both sides by 2 and you get this->

\(\large\frac{\cancel 2cos\theta}{\cancel 2}=\frac{\sqrt{3}}{2}\)
\(cos\theta=\large\frac{\sqrt{3}}{2}\)

now you got the value of \(cos\theta\) sooo just plug this value in the given expression \(2 \cos \theta + 1/ 2 \cos \theta - 1 \) to get its value

Answer 4

@imqwerty Oh, thank you!

Answer 5

np hokage (;

Answer 6

Lawl ;D

Answer 7

(/¯–‿･)/¯

Answer 8

omg, why did u delete that? xD

Answer 9

x'D cuz i ran outta chakra

Answer 10

my fav jitsu ;D

Answer 11

samee lel ;)

Answer 12

damn, qwerty lel

Answer 13

hahaha XD

Answer 14

ooooh ^^

Answer 15

:o ;)

Answer 16

tso deleted it </3

Answer 17

I'll repost it then ;p

qwerty is a qt ;)<3

Answer 18

so I got 4 root 3 + 1/ 4 root 3 - 1.. So now, I have to rationalize.. Yea? ;-;

Answer 19

Btw, @hxkage welcome to OpenStudy! :D

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Answer 20

@TheSmartOne Aww, thank ya! ;D

Answer 21

i think you made a little mistake there

\(2 \cos \theta + \large\frac{1}{ 2} \cos \theta - 1 \)

now substituting this value that we got- \(cos\theta=\large\frac{\sqrt{3}}{2}\)

\(\cancel 2 \times \large\frac{\sqrt{3}}{\cancel 2}+ \large\frac{1}{ 2}\times \large\frac{\sqrt{3}}{2}- 1 \)

Answer 22

Oh yea, sorry! I'm such a careless noob at times. Thanks tho. ;-; @imqwerty

Answer 23

its okay :) and np

Answer 24

Actually, it was:
\[\frac{ 2\cos \theta + 1 }{ 2\cos \theta - 1 }\]
sorry, my bad @imqwerty

Answer 25

So, I got:
\[\frac{ \sqrt{3}+1 }{ \sqrt{3} - 1}\]

Answer 26

oh okay
what you got is all correct and now you need to rationalize the denominator

Answer 27

alright, gimme a min ;-;

Answer 28

kkyz

Answer 29

I got (root 3 + 1) ^ 2 / 2

Answer 30

correct

Answer 31

Thanks, baruto! ;) Your patience is appreciated. @imqwerty

Answer 32

(; yw hokage

Answer 33

aha ;)