Posted by Anonymous on .
Prove:
cos4x = 8cos^4x  8cos^2x + 1
My Attempt:
RS:
= 4cos^2x (2cos^2x  1) + 1
= 4 cos^2x (cos2x) + 1
LS:
= cos2(2x)
= 2cos^2(2x)  1
= (cos^2(2))  cos^2(2x))  1

Prove:
8cos^4x = cos4x + 4cos2x + 3
My Attempt:
RS:
= cos2(2x) + 4cos(2x) + 3
= 2cos^2(2x)  1 + 2(4)cos^2(2x)  1 + 3

Math  Trig  Double Angles 
Michael,
When proving these, you should only change ONE side of the equation. Pick one side, and try to get it to match the other.
For the first one, it's important to note the doubleangle identity for cosines. There are three versions, but since both sides contain cosines, we're going to use the version that includes only cosines.
cos(2x) = 2cos^2(x)  1
cos(4x) = 8cos^4(x)  8cos^2(x) + 1
Since the RS seems like an expansion of the LS, we will work SOLELY with the LS to get it to match the RS.
cos(4x)
cos(2*2x) Doubleangle identity.
2cos^2(2x)  1 Doubleangle again.
2(cos2x * cos2x)  1 Doubleangle again.
2((2cos^2(x)  1)(2cos^2(x)  1))  1
Now, distribute the 2 into the first parentheses.
(4cos^2(x)  2)(2cos^2(x)  1)  1
Foil.
8cos^4(x)  4cos^2(x)  4cos^2(x) + 2  1
Simplify.
8cos^4(x)  8cos^2(x) + 1
That matches the RS, so we're done.
Try a similar process with your other proof. These do take a bit of practice. You may need to practice with more than the amount of homework that your teacher has assigned. It took me an extra 10 hours of additional practice in a week when we first learned these. Once you do enough, you'll see the patterns. It's crucial to memorize the identities, though. 
Math  Trig  Double Angles 
Anonymous,
Why is cos4x, cos(2*2x) instead of cos2(2x)? or do both mean the same thing?

Math  Trig  Double Angles 
paul,
verify identity
cos 2A sec A = 2cos A  sec A 
Math  Trig  Double Angles 
Michael,
cos2(2x) should be cos(2(2x)). That 2 can't be floating around outside the cosine function. In any case, cos(2(2x)) is the same as cos(2*2x).

Math  Trig  Double Angles 
Anonymous,
For the following lines,
2cos^2(2x)  1 Doubleangle again.
2(cos2x * cos2x)  1 Doubleangle again.
I don't get how you got the second line from the first line... 
Math  Trig  Double Angles 
Michael,
I was looking at another of your questions, and in Npgm's proof, he/she modified both sides of the equation. You absolutely cannot do this. You must change only one side of the equation. My teacher called this "math etiquette," but it is etiquette that must be followed. (I just wanted to stress this again.)

Math  Trig  Double Angles 
Anonymous,
Okay? But, I still don't get what you did in:
2cos^2(2x)  1 Doubleangle again.
2(cos2x * cos2x)  1 Doubleangle again. 
Math  Trig  Double Angles 
Michael,
I'll try to explain the two lines better...
2cos^2(2x)  1
This is like 2(cos^2(2x))  1.
In this line, we're changing only the cos^2(2x). When the squared is written on the "cos" part, that means the entire function is squared. Therefore, cos^2(2x) is the same as (cos(2x))^2.
When something is squared, you multiply it by itself. For example, 4^2 = 4*4, and (n+1)^2 = (n+1)(n+1).
Therefore, we are saying that (cos(2x))^2 = (cos(2x))(cos(2x)) = cos2x * cos2x. The rest of the line remained unchanged. 
Math  Trig  Double Angles 
Michael,
(Just to clarify, I wrote the oneside bit before I saw your other question. Sorry for the confusion.)

Math  Trig  Double Angles 
Anonymous,
Since you square the entire function, why is it (cos2x)(cos2x)? I mean, why don't you square the "2" into (cosx)(cos2x)?

Math  Trig  Double Angles 
Michael,
Do you mean... why isn't (cos(2x))^2 equal to cos(4x^2)?

Math  Trig  Double Angles 
Anonymous,
Ugh...nevermind I'm so confused...
If I did the right side, would it be more easy than the left side? I don't think I know how to expand with identities... 
Math  Trig  Double Angles 
Michael,
No, it would be much harder. It's always easier to expand than to condense.
Expansion of the identities is a very important concept.
cos(2x) = 2cos^2(x)  1
If it's cos(4x), then that's like cos(2*2x). Everything in the expansion is the same except the x will be 2x now, like 2cos^2(2x)  1.
All you're doing is repeating the expansion process until you can't do it anymore.
If you have more questions that are confusing you, then I can try to help. 
Math  Trig  Double Angles 
Anonymous,
Ah okay, I get it now, so "2x" is like "x", they are always together
I'm working on the second question now and I'm expanding the identity, but I'm kind of stuck...
8cos^4x = cos4x + 4cos2x + 3
LS:
= 8cos^4x
= 8(cos^2x)(cos^2x)
= 4(cos^4(2x))
= 4(cos^2(2x))*(cos^2(2x))
= 4(cos2x)*(cos2x)*(cos2x)*(cos2x)
= (4cos^2(2x))*(2cos^2(2x)  1)*(2cos^2(2x)  1) * (cos(2x)) 
Math  Trig  Double Angles 
Anonymous,
Made a mistake on the last line
= (4cos(2x))*(2cos^2(2x)  1)*(2cos^2(2x)  1) * (cos(2x)) 
Math  Trig  Double Angles 
Michael,
I'm working on it...

Math  Trig  Double Angles 
Michael,
Well, you've chosen the wrong side to work with. You cannot expand 8cos^4(x). For the doubleangle identity to work, a number has to be multiplied by the x. (The 4 isn't multiplied; it's the power.)
Work on the righthand side (cos4x + 4cos2x + 3) and see what you come up with. 
Math  Trig  Double Angles 
Anonymous,
RS:
= (cos2(2x)) + (4cos(2x)) + 3
= 2cos^2(2x)  1 + 8cos^2(2x)  1 + 3
= 2cos^2(2x) + 8cos^2(2x)  2 + 3
= 2cos^2(2x) + 8cos^2(2x) + 1
I don't think I did it right, but I can't think of any other way of doing it... 
Math  Trig  Double Angles 
Michael,
You didn't do 4cos(2x) correctly.
Our doubleangle identity is...
cos(2x) = 2cos^2(x)  1
(Memorizing it will be very helpful.)
4cos(2x)
4(2cos^2(x)  1)
Distribute the 4 in and continue. 
Math  Trig  Double Angles 
Anonymous,
And I still get stuck...
= cos2(2x) + 4(2cos^2(2x)1) + 3
= cos2(2x) + 8cos^2(2x)  4 + 3
= 2cos^2(2x)  1 + 8cos^2(2x)  4 + 3
= 2cos^2(2x) + 8cos^2(2x)  2
What am I doing wrong now? 
Math  Trig  Double Angles 
Michael,
When you evaluated the doubleangle (2x), you kept the 2x when you should have dropped the 2.
It should be...
cos(2(2x)) + 4(2cos^2(x)1) + 3
You are right that cos(2(2x)) = 2cos^2(2x)  1. 
Math  Trig  Double Angles 
Anonymous,
I give up...I can't solve this identity using the right side...
LS:
= 8(cos^2x)^2
= 8 (1+cos2x / 2)^2
= 8 (1 + 2cos2x + cos^2(2x) / 4)
= 2 + 4cos2x + 1 + cos(4x)
= cos(4x) + 4cos(2x) + 3 
Math  Trig  Double Angles 
Michael,
Well, you can't solve it with the left side... 8cos^4(x) does not equal 8(cos^2x)^2. It equals 8(cos^2(x))^2. The 2 is a power; it's not multiplied by the x, so you can't use the doubleangle identity.
You were on the right track before. If you want to make another attempt, I'll be happy to help. 
Math  Trig  Double Angles 
Anonymous,
Whenever I try to prove the identity with the right side, I keep getting different answer
RS:
= cos2(2x) + 4(2cos^2(x)1) + 3
= cos2(2x) + 8cos^2(x)  4 + 3
= 2cos^2(2x)  1 + 8cos^2(x)  4 + 3
= 2cos^2(2x) + 8cos^2(x)  2
...I think I'm making this too complicated than it should be... 
Math  Trig  Double Angles 
Anonymous,
additional to my answer:
= (2cos^2(x)1)*(2cos^2(x)1) + 8cos^2(x)  4 + 3
= 4cos^4(x) + 1 + 8cos^2(x)  4 + 3
= 4cos^4(x) + 8cos^2(x) 
Math  Trig  Double Angles 
Michael,
All right... your first post is entirely correct. It's not too complicated. My answer was about 10 steps.
(2cos^2(x)1)*(2cos^2(x)1) + 8cos^2(x)  4 + 3 is where you went a bit wrong.
2cos^2(2x) + 8cos^2(x)  2
2((2cos^2(x)1)*(2cos^2(x)1)) + 8cos^2(x)  2
Actually, all you did was forget to bring your 2 in front of cos^2(2x) down to the next line.
Now, distribute the 4 in and foil those two parentheses. 
Math  Trig  Double Angles 
Michael,
Distribute the 2 in.*

Math  Trig  Double Angles 
Anonymous,
And I get:
= 8cos^4(x) + 8cos^2(x)
I don't know what to do now, how do I eliminate 8cos^2(x)? 
Math  Trig  Double Angles 
Anonymous,
And where did you get the " 2" from 2cos^2(2x) + 8cos^2(x)  2

Math  Trig  Double Angles 
Michael,
2((2cos^2(x)1)*(2cos^2(x)1)) + 8cos^2(x)  2
Distribute in the 2.
(4cos^2(x)2)(2cos^2(x)1) + 8cos^2(x)  2
Foil.
8cos^4(x)  4cos^2(x)  4cos^2(x) + 2 + 8cos^2(x)  2
Simplify. (The 2's cancel.)
8cos^4(x)  8cos^2(x) + 8cos^2(x)
Simplify. (The 8cos^2(x)'s cancel.)
8cos^4(x)
Questions? 
Math  Trig  Double Angles 
Michael,
To answer your other question, the "2cos^2(2x) + 8cos^2(x)  2" comes from the previous line...
2cos^2(2x)  1 + 4(2cos^2(x)  1) + 3
Just the consonants: 1  4 + 3 = 2 
Math  Trig  Double Angles 
Anonymous,
Foil.
8cos^4(x)  4cos^2(x)  4cos^2(x) + 2 + 8cos^2(x)  2
^^ where did you get 4cos^2(x)  4cos^2(x)?
When I try to prove the identity, I get:
= 8cos^2(x)  2  4cos^2(x) + 8cos^2(x)  2 
Math  Trig  Double Angles 
Anonymous,
= 8cos^4(x)  2  4cos^2(x) + 8cos^2(x)  2**** I mean

Math  Trig  Double Angles 
Michael,
All these numbers and things are getting confusing. Let's try to simplify this.
Could you tell me how you foiled this?
(4cos^2(x)2)(2cos^2(x)1) 
Math  Trig  Double Angles 
Anonymous,
= (4cos^2(x)2)(2cos^2(x)1)
= (8cos^4(x) + 2) (4cos^2(x)+2) 
Math  Trig  Double Angles 
Michael,
When you foil, you shouldn't have any parentheses, so this is what you had (minus the parentheses)...
8cos^4(x) + 2  4cos^2(x) + 2
First good
Outer no
Inner good
Last good
The outer is 4cos^2(x) multiplied by 1, so it should be 4cos^2(x).
(4cos^2(x)2)(2cos^2(x)1)
8cos^4(x)  4cos^2(x)  4cos^2(x) + 2
8cos^4(x)  8cos^2(x) + 2 
Math  Trig  Double Angles 
Anonymous,
thanks, I get it now! I know what I did wrong, I didn't distribute the parenthesis correctly...
thanks for putting up with me :)
I'll definitely keep practicing until I get this right away 
Math  Trig  Double Angles 
Michael,
Glad to help. Good luck! :)