# Homework Help: Trig Identity RHS LHS problem

1. Nov 6, 2007

### rum2563

[SOLVED] Trig Identity

1. The problem statement, all variables and given/known data
cos^4 (x) = (3/8) + (1/2)(cos(2x)) + (1/8)(cos(4x))

2. Relevant equations
cos2x = 2cos^2 x - 1

cos^2 x = 1 - sin^2 x

3. The attempt at a solution

Can someone please give me hints? Thanks.

2. Nov 6, 2007

### rock.freak667

Try proving the RHS to the LHS

and use the formula cos2A=cos$^2$A-sin$^2$A=2cos$^2$A-1=1-2sin$^2$A

3. Nov 6, 2007

### rum2563

ok. Here is what I did:

= (3/8) + (1/2)(cos(2x)) + (1/8)(cos(4x))
= (3/8) + (2cos^2 x - 1) + (1/4cos^2 2x - 1)
= (3/8) + (2cos^2 x - 1) + (1/2cos^4 x - 1 - 1)

This is where I don't know what to do. Any help would be great. Thanks.

4. Nov 6, 2007

### Avodyne

You have one mistake going from the 1st line to the 2nd, and another going from the 2nd to the 3rd.

5. Nov 6, 2007

### rum2563

I am sorry. I didn't realize that. Here is my new try:

= (3/8) + (1/2)(cos(2(x)) + (1/8)(cos(4(x))
= (3/8) + (1/2)(2cos^2 x - 1) + (1/8)(2cos^2 2x - 1)
= (3/8) + (2cos^2 x) - (1/2) + (1/4cos^2 2x) - (1/8)

Does this seem right? Please help. I am running out of time. Thanks.

6. Nov 6, 2007

### Avodyne

You still have a mistake going from the 2nd to the 3rd line.

And, you eliminated some (cos 2x)'s, but you still have one in your last line.

7. Nov 6, 2007

### rum2563

I am getting confused now. Can you elaborate a bit more please?

Is this rite?

= (3/8) + (((2cos^2 x)-1)/2) + (((2cos^2 2x)-1)/2)
= (3/8) + (cos^2 x) - (1/2) + (cos ^2 2x)/4 - (1/8)

I have been trying this for a long time. I think I am on the right track. Please help. Thanks.

Side note: I read something about power reduction formulae on the internet, and our teacher hasn't taught us that. So we can only use trig identities to solve this question.

Last edited: Nov 6, 2007
8. Nov 6, 2007

### Avodyne

Yes, that's right, and you are one the right track.

Now, you still have a cos 2x in your next-to-last term in the last line. You want to write everything in terms of cos x.

9. Nov 6, 2007

### rum2563

Thanks. Actually, that's the part which is most confusing.

Here is what I make of it:

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

Is this right? I know that cos2x = 2cos^2 x - 1. But I don't know whether the exponent would change or not. Please help. Thanks.

10. Nov 6, 2007

### Avodyne

No, it's not right. Suppose you have y^2/4, and I told you that y = 2 z^2 - 1; can you express y^2/4 in terms of z?

11. Nov 6, 2007

### rum2563

I am so sorry. Let me try again:

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

Is this right? I hope it is because this time I really thought about it.

12. Nov 6, 2007

### Avodyne

Yes, it's right!

Now, with c = cos x, you have

(3/8) + c^2 - (1/2) + ((2c^2 - 1)^2)/4 - (1/8)

Expand out the square, and simplify as much as possible.

13. Nov 6, 2007

### rum2563

Wow. Thanks very much Avodyne. Although it took me a long time to understand, it was still worth it.

I am going to sleep now, but when I expanded the equation ((2c^2 - 1)^2)/4 , I got c^4-c^2+1/4 which was the key to solving this question. (And Thank God I got that, lol)

I am very thankful to Avodyne(especially), and off course to rock.freak667 in helping me do this question and understand the concept. You are the best guys.

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