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October 24, 2014

October 24, 2014

Posted by **Anonymous** on Saturday, November 17, 2007 at 6:17pm.

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**, Saturday, November 17, 2007 at 8:34pmWhen 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 double-angle 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) Double-angle identity.

2cos^2(2x) - 1 Double-angle again.

2(cos2x * cos2x) - 1 Double-angle 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**, Saturday, November 17, 2007 at 8:39pmWhy is cos4x, cos(2*2x) instead of cos2(2x)? or do both mean the same thing?

- Math - Trig - Double Angles -
**paul**, Saturday, March 22, 2008 at 3:24pmverify identity

cos 2A sec A = 2cos A - sec A

- Math - Trig - Double Angles -
- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 8:44pmcos2(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**, Saturday, November 17, 2007 at 8:47pmFor the following lines,

2cos^2(2x) - 1 Double-angle again.

2(cos2x * cos2x) - 1 Double-angle again.

I don't get how you got the second line from the first line...

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 8:48pmI 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**, Saturday, November 17, 2007 at 8:51pmOkay? But, I still don't get what you did in:

2cos^2(2x) - 1 Double-angle again.

2(cos2x * cos2x) - 1 Double-angle again.

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 8:53pmI'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**, Saturday, November 17, 2007 at 8:54pm(Just to clarify, I wrote the one-side bit before I saw your other question. Sorry for the confusion.)

- Math - Trig - Double Angles -
**Anonymous**, Saturday, November 17, 2007 at 8:59pmSince 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**, Saturday, November 17, 2007 at 9:01pmDo you mean... why isn't (cos(2x))^2 equal to cos(4x^2)?

- Math - Trig - Double Angles -
**Anonymous**, Saturday, November 17, 2007 at 9:03pmUgh...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**, Saturday, November 17, 2007 at 9:07pmNo, 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**, Saturday, November 17, 2007 at 9:12pmAh 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**, Saturday, November 17, 2007 at 9:13pmMade a mistake on the last line

= (4cos(2x))*(2cos^2(2x) - 1)*(2cos^2(2x) - 1) * (cos(2x))

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 9:16pmI'm working on it...

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 9:21pmWell, you've chosen the wrong side to work with. You cannot expand 8cos^4(x). For the double-angle identity to work, a number has to be multiplied by the x. (The 4 isn't multiplied; it's the power.)

Work on the right-hand side (cos4x + 4cos2x + 3) and see what you come up with.

- Math - Trig - Double Angles -
**Anonymous**, Saturday, November 17, 2007 at 9:30pmRS:

= (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**, Saturday, November 17, 2007 at 9:33pmYou didn't do 4cos(2x) correctly.

Our double-angle 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**, Saturday, November 17, 2007 at 9:43pmAnd 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**, Saturday, November 17, 2007 at 9:51pmWhen you evaluated the double-angle (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**, Saturday, November 17, 2007 at 10:01pmI 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**, Saturday, November 17, 2007 at 10:06pmWell, 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 double-angle 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**, Saturday, November 17, 2007 at 10:10pmWhenever 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**, Saturday, November 17, 2007 at 10:24pmadditional 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**, Saturday, November 17, 2007 at 10:38pmAll 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**, Saturday, November 17, 2007 at 10:39pmDistribute the 2 in.*

- Math - Trig - Double Angles -
**Anonymous**, Saturday, November 17, 2007 at 10:50pmAnd 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**, Saturday, November 17, 2007 at 10:52pmAnd where did you get the "- 2" from 2cos^2(2x) + 8cos^2(x) - 2

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 10:57pm2((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**, Saturday, November 17, 2007 at 11:02pmTo 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**, Saturday, November 17, 2007 at 11:08pmFoil.

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**, Saturday, November 17, 2007 at 11:09pm= 8cos^4(x) - 2 - 4cos^2(x) + 8cos^2(x) - 2**** I mean

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 11:10pmAll 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**, Saturday, November 17, 2007 at 11:12pm= (4cos^2(x)-2)(2cos^2(x)-1)

= (8cos^4(x) + 2) (-4cos^2(x)+2)

- Math - Trig - Double Angles -
**Michael**, Saturday, November 17, 2007 at 11:18pmWhen 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**, Saturday, November 17, 2007 at 11:26pmthanks, 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**, Saturday, November 17, 2007 at 11:28pmGlad to help. Good luck! :)

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