polynomial question (1 Viewer)

[stainmepink]

hey guys help me with this question my test is tomorrow

part a) use de moivres theorem to express cos4ø and sin4ø in terms of cosø and sin ø and hence find cot4ø in terms of cotø

ive done that

part b) explain why cotø is a solution to x^4 - 6x^2 +1=0
how do you explain it?

 

[stainmepink]

the answer to the first part is




(cotø)^4 - 6(cotø)^2 +1
----------------------------------
4(cotø)^3 - 4(cotø)

its the second part i don't understand

 

[pLuvia]

If you let
(cot⁴ø-6cot²ø+1)/(4cot³ø-4cotø)=0
(cot⁴ø-6cot²ø+1)=0
Let x=cotø
x⁴-6x²+1=0
Hence cotø is a solution to that equation