Question

Evaluate :
\(\cos 12 \cos 24 \cos 36 \cos 48 \cos 72 \cos 84\)

Answer 1

@Mertsj

Answer 2

I was thinking to simplify \(\cos 72\) more as the pattern was going +12 ... but 48 + 12 = 60 that broke the pattern.

Answer 3

\(\cos 12 \cos (2(12)) \cos (3(12)) \cos (4(12)) \cos (6(12)) \cos (7(12))\)

Answer 4

We have multiple angles.

Answer 5

\( (\cos 12 \cos 24 \cos 48 )( \cos 36 \cos 72 \cos (180-96)\)

Oh I have something useful here.

Answer 6

cosx .cos(60-x) . cos(60+x) = cos3x/4

Answer 7

use this here

Answer 8

how about noticing
12+48=60, 12-48=-36
24+36=60, 24-36=-12

Answer 9

\( - (\cos 12 \cos 48 \cos 24 \cos 96) (\cos 36 \cos 72) \)

Hmn .. @electrokid but what would I get from there?

Answer 10

my bad, its 72,

Answer 11

\[\cos(12°) \cos(24°) \cos(36°) \cos(48°) \cos(72°) \cos(84°)\]
\[1/64\]

Answer 12

Maybe double angle for \(\cos\) would be good.

Answer 13

\[
\cos12\cos48={1\over2}[\cos(12-48)-\cos(12+48)]={1\over2}\left(\cos36-{1\over2}\right)={1\over2}\cos36-{1\over4}
\]

Answer 14

hmmmm...

Answer 15

\[
\cos24\cos36={1\over2}[\cos(24-36)-\cos(24+36)]={1\over2}\left(\cos12-{1\over2}\right)={1\over2}\cos12-{1\over4}
\]

Answer 16

\[
\cos(2x) = \cos^2(x)-\sin^2(x)
\]

Answer 17

@electrokid, it'd be nice if there were more factors of \(\cos(5(12))\) to work with... Anything we can do with the remaining \(\cos(6(12))\) and \(\cos(7(12))\)?

Answer 18

I hope the product expansion of these parantheses would provide us with something
\[\cos (x)\cos (2x)\cos(3x)\cos(4x)\cos(6x)\cos(7x)\]

Answer 19

let θ=mean of the angles?

Answer 20

\[\huge\frac{ 1 }{ 4 }sin60*\cos12*\cos24\]

Answer 21

@mathslover what is the answer??

Answer 22

I am getting it as \(\cfrac{1}{64}\)

Answer 23

hmmm i got that too xD

Answer 24

is there cos 60 also in the question??

Answer 25

no!

Answer 26

then how come u got 1/64 ??

Answer 27

oh yes yes

Answer 28

the answer is \[\huge\frac{ 1 }{ 64 }\]

Answer 29

\[\frac{ 1 }{ 64 } \approx 0.0156250\]

Answer 30

the answer is 1/64

Answer 31

i took cos 60 also so i got my previous answer!!

Answer 32

cos(12^(o))cos(24^(o))cos(36^(o))cos(48^(o))cos(72^(o))cos(84^(o))

Take the cosine of 12^(o) to get 0.9782.
(0.9782)(cos(24^(o))cos(36^(o))cos(48^(o))cos(72^(o))cos(84^(o)))

Take the cosine of 24^(o) to get 0.9136.
(0.9782)(0.9136)(cos(36^(o))cos(48^(o))cos(72^(o))cos(84^(o)))

Take the cosine of 36^(o) to get 0.809.
(0.9782)(0.9136)(0.809)(cos(48^(o))cos(72^(o))cos(84^(o)))

Take the cosine of 48^(o) to get 0.6691.
(0.9782)(0.9136)(0.809)(0.6691)(cos(72^(o))cos(84^(o)))

Take the cosine of 72^(o) to get 0.309.
(0.9782)(0.9136)(0.809)(0.6691)(0.309)(cos(84^(o)))

Take the cosine of 84^(o) to get 0.1045.
(0.9782)(0.9136)(0.809)(0.6691)(0.309)(0.1045)

Multiply 0.9782 by 0.9136 to get 0.8936.
(0.8936)(0.809)(0.6691)(0.309)(0.1045)

Multiply 0.8936 by 0.809 to get 0.7229.
(0.7229)(0.6691)(0.309)(0.1045)

Multiply 0.7229 by 0.6691 to get 0.4837.
(0.4837)(0.309)(0.1045)

Multiply 0.4837 by 0.309 to get 0.1495.
(0.1495)(0.1045)

Multiply 0.1495 by 0.1045 to get 0.0156.
(0.0156)

Remove the parentheses around the expression 0.0156.
0.0156


So this is how I got it which i got 0.0156
and 1/64 is ~ 0.0156

DO YOU UNDERSTAND HOW TO DO IT NOW, EVERYONE?

Answer 33

@some_someone took from .. software? lol

Answer 34

yeah i checked it using wolfram?

Answer 35

:)

Answer 36

Yeah! Its \(\cfrac{1}{64}\) . You're right @mathslover