Something obvious?

[Dancer25]

Any ideas on how to easily compute cos(12)cos(24)cos(48)cos(96)? I tried product to sum but it got messy

 

[galactus]

I assume this has to be done without a calculator?.

$\displaystyle \prod_{n=0}^{3}[cos(12\cdot 2^{n})]=\frac{-1}{16}$

Are there guidelines you must follow?. I am asking because there are various ways to go about it.

Note that they're all multiples of 12.

 

[Dancer25]

I can't use a calculator; no other guidelines.

 

[mmm4444bot]

cos(12) cos(24) cos(48) cos(96) This notation states the angles in radians.

cos(12°) cos(24°) cos(48°) cos(96°) This notation states the angles in degrees.

I'm wondering if this exercise's angles are supposed to be in degrees.

 

[Deleted member 4993]

Any ideas on how to easily compute cos(12)cos(24)cos(48)cos(96)? I tried product to sum but it got messy

Assuming you are talking degrees:

cos(12)cos(24)cos(48)cos(96)

= 1/[2sin(12)][2sin(12)cos(12)]cos(24)cos(48)cos(96)

= 1/[2sin(12)][sin(24)]cos(24)cos(48)cos(96)

= 1/[4sin(12)][2sin(24)cos(24)]cos(48)cos(96)

= 1/[4sin(12)][sin(48)]cos(48)cos(96)

= 1/[8sin(12)][2sin(48)cos(48)]cos(96)

= 1/[8sin(12)][sin(96)]cos(96)

= 1/[16sin(12)][2sin(96)cos(96)]

= 1/[16sin(12)][sin(192)]

= 1/[16sin(12)][sin(180+12)]

= 1/[16sin(12)][ - sin(12)]

= - 1/16

After BigGlenn declared that he could not proceed - I surmised that the steps must not be as obvious as I had thought. So I completed the work .....

 

[BigGlenntheHeavy]

$\displaystyle If \ there \ is \ something \ obvious, \ without \ a \ calculator, \ I \ missed \ it.$

 

[Deleted member 4993]

Glenn,

I have completed the steps above....

 

[BigGlenntheHeavy]

\(\displaystyle Subhotosh \ Khan, \ good \ show, \as \ I \ followed \ your \ analysis \ to \ the \ end, \ however\)

$\displaystyle was \ it \ obvious, \ perhaps \ so. \ Anyways, \ good \ show, \ I'll \ remember \ this \ one.$

$\displaystyle With \ trig. \ identities, \ one \ can \ do \ wonders.$

$\displaystyle Note: \ It \ is \ Glenn, \ not \ Glen.$

 

[Deleted member 4993]

$\displaystyle was \ it \ obvious, \ perhaps \ so. \ Anyways, \ good \ show, \ I'll \ remember \ this \ one.$ ...Obvious??!! No - like almost all other trig-identities - I stumbled into it. And that is why I am afraid I won't remember it next time - just have to redo the stumbling around...$\displaystyle With \ trig. \ identities, \ one \ can \ do \ wonders.$

$\displaystyle Note: \ It \ is \ Glenn, \ not \ Glen.$ ... Corrected