Question

Find the exact value of cos(7pi/12) using half-angle identities. I'm still not sure how to solve this :( Any help is appreciated!

Answer 1

see first of all think it dis way, ( i dont have calculator, but still i have to do d question, how can it b done)

Answer 2

well I know I'm supposed to pick "known" angles from the Unit Circle around where (7pi/12) would be or something like that

Answer 3

But I don't know how to do that without picking the wrong thing

Answer 4

yep, now u know the values of cos of pi/6, pi/3,pi,4,pi/2

Answer 5

How can you tell which one it is that you can use?

Answer 6

Hint: Do you know the value of cos(2*7pi/12) i.e. the value of cos(14pi/12). See if you can take it off from here

Answer 7

sum of the denominators of which two angels( out of the four i typed) is equal to 7, nd there product is 12?

Answer 8

pi/3 and pi/4? (:

Answer 9

u have got it :)

Answer 10

thats it, now apply the addition formula for cos

Answer 11

You mean cos = theta/2 = +- square root 1 + cos theta / 2?

Answer 12

I thought the question meant using half angle formulae, hence the hint about cos(14pi/12).
Otherwise pi/3 and pi/4 are perfectly fine :)

Answer 13

Oh okay (:

Answer 14

Or do you mean cos (A + B) = cosAcosB - sinAsinB

Answer 15

nah, i meant cos(a+b)= cosacosb-sinasinb

Answer 16

I'm sorry, there's just so many formulas being tossed around today in my class lol.

Answer 17

Oh alrighty. What about sinA and sinB?

Answer 18

we know that values also, ( a gud saying, if u knw the one, u knw d rest, so if u knw the value of cosx, u can find the values of all other trignometric functions of x)

Answer 19

if you have to use half angle formullas only,,use these :
1)cos x = 2cos^2 (x/2) -1 or any other equivalent form.
2) your x here is 7pi/12
3)cos (7pi/6) = cos(pi + pi/6) = -(cos pi/6)
try now..

Answer 20

@shubhamsrg would -(cos pi/6) be \[\sqrt{3} / 2\]

Answer 21

i am sorry above..x is 7pi/6 and not 7pi/12

Answer 22

yep its that only

Answer 23

it'd be -sqrt(3)/2

Answer 24

and i owe u a big apology, i wasted a lot of ur time, i didnt read the question well, u asked to use half angle iddentities, nd i used additional iddentity, sorry

Answer 25

Oh please, don't be (: You helped me either way! I really appreciate it (:

Answer 26

If through this though, we got -sqrt3/2, do we plug that into cos theta / 2 = sqrt 1 + cos theta / 2

Answer 27

\[\cos \theta = \pm \sqrt{1 + \cos \theta} / 2\]

Answer 28

didnt get your query.,,come again ..

Answer 29

ty:)

Answer 30

Do we continue solving? Because I don't see \[-\sqrt{3} / 2\] as an answer (:

Answer 31

And sure thing @dg123 (:

Answer 32

well i'll put an eqn,,you try afterwards :
cos (7pi/6) = -cos(pi/6) = 2cos^2 (pi/12) -1
solve for cos(pi/12) from here..

Answer 33

hope its clear how i got the eqn.

Answer 34

Okay, give me a second (:

Answer 35

Yeah I'm stuck :/ Lol I'm sorry.

Answer 36

@FoolAroundMath we started with pi/3 and pi/4, may you carry off from there please? If not, I understand (: I just got a little lost here.

Answer 37

hmmm.. :| :P

Answer 38

When I solved the equation you gave me @shubhamsrg I got -0.86

Answer 39

1 min..

Answer 40

Alrighty

Answer 41

well i get cos(7pi/12) as 0.26 approx

Answer 42

hmmm... I wonder where I messed up :p This is how I did it... hold on please

Answer 43

dont convert into fraction form,,leave it in rational form,,afterall you'll have to match with the options as well..

Answer 44

Ooh, that's my problem than :/ I don't know how to do that without turning it into like a decimal and stuff

Answer 45

There are essentially two methods to solve this, both use the same trigonometric formula :
cos(A+B) = cosAcosB - sinAsinB

1. take A = pi/4 and B = pi/3 and just plug in. (Assuming you know sin/cos of pi/4 and pi/3)

2. This uses half-angle formula or twice angle formula and I will demonstrate that they are essentially the same as the first equation mentioned.

If B = A, then plugging it in the cos(A+B) formula, we get
cos(A+A) = cosAcosA - sinAsinA
\[\cos(2A) = \cos^{2}A - \sin^{2}A\]
\[\cos(2A) = \cos^{2}A - (1-\cos^{2}A)\] \[(\cos^{2}\theta + \sin^{2}\theta = 1)\]
Thus,
\[\cos(2\theta) = 2\cos^{2}\theta - 1\]
or
\[\cos(\theta/2) = \pm \frac{\sqrt{1+\cos\theta}}{2}\]
So, you need cos(7pi/12)
Lets see if you know the value of cos(14pi/12).

Answer 46

alright,,leme try to put up
let cos (pi/12)=x
so you have
-sqrt(3)/2 = 2x^2 -1
(1 - sqrt(3)/2 )/2 = x^2

x = sqrt (2- sqrt(3)) /2

so this is your ans

note that we ignored the -ve sqrt here..

Answer 47

(14pi/12) umm... 7pi/6? <:)

Answer 48

cos(14pi/12) = cos(pi + pi/6) = -cos(pi/6)
\[\cos\frac{14\pi}{12} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}\]
So using the half-angle formula that we derived above from the parent formula
cos(A+B) = cosAcosB - sinAsinB,
we can plug it in to get the value of cos(7pi/12)

Answer 49

@shubhamsrg what did we ignore?

Answer 50

???

Answer 51

You said at the end "note that we ignored the -ve sqrt here..."

Answer 52

i mean x will have 2 values according to our solution,,but we have to ignore the -ve solution since cos(7pi/12) must be positive

Answer 53

Oooh, okay. gotcha

Answer 54

@FoolAroundMath Thank you for the first part of the explanation, that cleared up a lot (: Your second part added onto what Shubhamsrg was saying (thank you by the way) and made it easier to understand (:

Answer 55

Thank you @shubhamsrg I know I'm not easy to work with Lol I appreciate your time spent helping me

Answer 56

glad to help ^_^

Answer 57

@FoolAroundMath did a lot of hard work(writing)..he deserves more of the credit thus :)

Answer 58

I wish I could give credit to everyone Lol

Answer 59

Like the best answer thingy Lol

Answer 60

Nah, I have experience with Latex, so typing equations is pretty fast .. And I am bored at the moment, so why not xD.
Glad I could help :)

Answer 61

Lol! (:

Answer 62

and i'dl like to make a correction,,we dont have to ignore the -ve part..but we have to ignore the +ve root..
so ans is the -ve of what was earlier..

Answer 63

cos(7pi/12) has to be -ve,,sorry,,just noticed..hmm

Answer 64

Oh, okay. So would it be (- sqrt. 2 - sqrt(3) all divided by 2)

Answer 65

yep..the very same :)

Answer 66

Okay, awesome (: How do you tell when to make it negative or positive?

Answer 67

7pi/12 = 105 degrees.. cos is -ve for that definitely

Answer 68

\[\cos\theta\] is positive in first and fourth quadrant and negative in others

Answer 69

Oh okay, because it's in the 2nd quadrant?

Answer 70

Ah, right (: