Comments (7) on “Find an identity for cos(4t) in terms of cos(t). i have used 2cos^2(2t)-1 but i need to rewrite that”
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1 Cos (pi/2)=0 Cos (pi) =-1 Cos (3pi/2)=0 Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period). But you can always reduce it. Cos (0)= 1 Cos (4pi/2) = cos (2pi)=1 Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1) etc... the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically. You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t) Sin^2 (t)=-Cos^2 (t)+1 (all A^2+B^2=C^2) Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1) Cos (2×2t)= 2Cos^2 (2t) - 1
The question in the problem wants to calculate the identity for cos(4t) in terms of cos(t) and base on the given and further computation, I would say that the answer would be 2cos^2(2t)-1. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1 (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
The question in the problem wants to calculate the identity for cos(4t) in terms of cos(t) and base on the given and further computation, I would say that the answer would be 2cos^2(2t)-1. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
I'm having the same problem
[tex]\cos4t=2\cos^22t-1=2(2\cos^2t-1)^2-1[/tex]
[tex]\cos4t=8\cos^4t-8\cos^2t+1[/tex]
Use double angle formula for cosine:
[tex]cos (2a) = 2cos^2(a) - 1[/tex]
Apply to cos(4t)
[tex]cos (2(2t)) = 2 cos^2 (2t) - 1 \\ cos (2t) = 2 cos^2 t - 1[/tex]
Sub 2nd equation into 1st equation:
[tex]cos(4t) = 2 (2 cos^2 t -1)^2 - 1 \\ = 8 cos^4 t -8cos^2 t +1[/tex]
Hello here is a solution :
[tex]Find an identity for cos(4t) in terms of cos(t)[/tex]
Answer
Find out the cos(4t) in terms of cos(t) .
To prove
As given the identity in the question be cos(4t) .
It is written as
[tex]cos(4t) = cos(2(2t))[/tex]
Now using the trignometric formula
[tex]cos2A = 2cos^{2}A - 1[/tex]
[tex]cos2(2t) = 2(2cos^{2}t -1)^{2} -1[/tex]
Apply (a +b )² = a² + b² + 2ab
[tex]cos2(2t) = 2(4cos^{4}t - 4cos^{2}t +1) -1[/tex]
Simplify the above
[tex]cos2(2t) = 8cos^{4}t - 8cos^{2}t +1[/tex]
Therefore the expression cos(4t) in terms of cos(t) is
[tex]cos2(2t) = 8cos^{4}t - 8cos^{2}t +1[/tex]