Evaluating cos(pi/12) |

[@georgiageo1681]

Another way: do cos(45-30) and use formula cos(a-b)=cos a cos b+sin a sin b

[@neuralwarp]

6:12 You didn't explain your logic. The base of the big triangle obviously has a 180° point, so deducting 30° from the small triangle means the obtuse angle of the isoceles is 150°. The internal angles sum to 180° so that is shared as 150° + 2×15°.

[@moritzalshuth7239]

The 4th way: Use 4 cos^3(phi) = 3 cos(phi) + cos(3 phi). Set y = cos(pi/12) and note that cos(3 pi/12) = cos(pi/4) = sqrt(2)/2. Solve cubic eqn. 4 y^3 - 3 y - sqrt(2)/2 = 0. Try rational solns adjoined {sqrt(2)} by subst. u := sqrt(2) y. Now solve 2 u^3 - 3 u - 1 = 0. Easily guess u = -1 as the (negative) rational solution. However, cos(pi/12) is positive, so we further need to find max{solns}, e.g. via long division: (1 + sqrt(3))/2. Quite tedious but after backsubstution we can confirm that cos(pi/12) = (sqrt(2) + sqrt(6)) / 4.

[@parthacreation1117]

Cos 15° = cos (45-30)= cos45 cos30+sin45 sin30= 1/√2 .√3/2 + 1/√2. 1/2= (√3+1)/2√2 ( ans )

[@robertveith6383]

The degree symbol needs to be next to each of the 45 and 30 numbers. 2sqrt(2) in the denominator needs to be inside of parentheses.

[@zehradiyab3439]

4th method
In 30°-60°-90° triangle ABC angle B is a right angle, if AD is angle bisector for the angleA is equal 30° and angle BAD=15°
AB/AC=BD/CD
sqrt(3)/2=x/1-x;
AB=sqrt(3), AC=2, BD=x, DC=BC-BD=1-x
Then x=sqrt(3)/[2+sqrt(3)]
AD²=3+[3/(2+sqrt(3))²]=...
=6[1+sqrt(3)]²/[2+sqrt(3)]²
AD=sqrt(6)[1+sqrt(3)]/[2+sqrt(3)]
cos15°=AB/AD=sqrt(3)÷[sqrt(6)[1+sqrt(3)]/[2+sqrt(3)]=...
=[aqrt(2)+sqrt(6)]/4

[@robertveith6383]

* x/(1 - x)

[@scottleung9587]

I used the half-angle formula, but wasn't quite sure how to express it in simplest terms.

[@mbmillermo]

interesting that it comes out to ½√2(½√3+½) - that's the three most famous sin and cos numbers put together.

[@SyberMath]

That's indeed interesting

[@yurenchu]

Indeed quite remarkable. But I've come to the realization that It's directly due to the fact that
cos(15°) = cos(45°-30°)
= cos(45°)cos(-30°) - sin(-30°)sin(45°)
= cos(45°)cos(30°) + sin(45°)sin(30°)
... cos(45°) = sin(45°) = ½√2 , cos(30°) = ½√3 , and sin(30°) = ½ ...
= (½√2)(½√3) + (½√2)(½)
= (½√2)(½√3 + ½)
I expect we'll see the same in cos(75°) .

[@mbmillermo]

@@yurenchu -- Thanks! Nice work. I was thinking there must be some underlying reason, and you just showed me what it was.

[@yurenchu]

@@mbmillermo You're welcome!

[@GreyJaguar725]

the best way: calculator method

[@kianmath71]

(Sqrt(2) + sqrt(6)) / 4

[@mikehood6745]

I'm going to have to take away that trick at 4:29 and give it some thought!

[@rakenzarnsworld2]

Answer: 0.9659

[@stevemonkey6666]

It would be way easier just to put the little degree symbols in rather than talk about it all the time. There is another method which is the easiest method and which is the method that almost everyone is going to use which is....... Use a calculator 😁

[@giuseppemalaguti435]

??? cos15=√(1+cos30)/2... perché un problema così semplice?

[@CriticSimon]

Dude, it’s not simple for everyone 😁