find the value of cos20
Answers
Answered by
12
Cos[60] = 4 Cos[20]^3 - 3 Cos[20]
1/2 = 4 Cos[20]^3 - 3 Cos[20]
Hence Cos[20] is the real root of the irreducible cubic
x^3 - 3/4 x - 1/8 = 0
This has the real root:
r = (a^(1/3) + b^(1/3))/2
where a = (1+Sqrt[-3])/2, b = (1-Sqrt[-3])/2 are complex conjugates
1/2 = 4 Cos[20]^3 - 3 Cos[20]
Hence Cos[20] is the real root of the irreducible cubic
x^3 - 3/4 x - 1/8 = 0
This has the real root:
r = (a^(1/3) + b^(1/3))/2
where a = (1+Sqrt[-3])/2, b = (1-Sqrt[-3])/2 are complex conjugates
Answered by
1
Answer:
Step-by-step explanation:
Decimal Form:
0.93969262
Similar questions