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).

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

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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 in terms of cos(2t).

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]