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Find an identity for cos(4t) in terms of cos(t). i have used 2cos^2(2t)-1 but i need to rewrite that

Posted on October 23, 2021 By Emmanuellugo40 7 Comments 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

Find an identity for cos(4t) in terms of cos(t).
i have used 2cos^2(2t)-1 but i need to rewrite that in terms of cos(2t).

Mathematics

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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”

  1. lovenot1977 says:
    October 23, 2021 at 10:31 am

    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

    Reply
  2. wolfiewolffromsketch says:
    October 23, 2021 at 11:53 am

    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

    Reply
  3. lilblakey69 says:
    October 23, 2021 at 1:07 pm

    I'm having the same problem

    Reply
  4. calebcoolbeans6691 says:
    October 23, 2021 at 2:30 pm

    [tex]\cos4t=2\cos^22t-1=2(2\cos^2t-1)^2-1[/tex]
    [tex]\cos4t=8\cos^4t-8\cos^2t+1[/tex]

    Reply
  5. nicolemaefahey says:
    October 23, 2021 at 9:48 pm

    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]

    Reply
  6. aorilneedshelp6636 says:
    October 24, 2021 at 6:32 am

    Hello here is a solution :
    [tex]Find an identity for cos(4t) in terms of cos(t)[/tex]

    Reply
  7. kezin says:
    October 24, 2021 at 6:56 am

    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]

    Reply

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