Help Me Verify This Identity (Math)

[Moto]

I'm doing double angle identities in math right now and I have a problem that I cannot solve. I'm being asked to verify the following identity:

I drew it to make it easier on the eyes, but if you want to see it typed out, it's:

8((sin^2)x)((cos^2)x) = 1-cos4x

Thanks for any help.

 

[swan_song1973]

we're dealing with sines and shit in my class too, it sucks doesnt it?

 

[Xendance]

Americans saying 'Math' really does piss me off.

 

[barbarbarbarbar]

damn i'm still at functions

 

[Moto]

Americans saying 'Math' really does piss me off.

Fine, "Maths."

Please just ignore this thread if all you have to say is how pissed off our differences make you.

 

[Xendance]

Fine, "Maths."

Please just ignore this thread if all you have to say is how pissed off our differences make you.

It was a joke. Chillax.

 

[Moto]

Sorry, I'm just way frustrated because I've been staring at this problem and a couple others for hours and can't figure them out. So I have very little patience right now.

 

[Dan]

I pulled a few trig identities off of Wikipedia:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities
All you really need is the double angle formulas and the Pythagorean identity, or just the power reduction formula which is what this question is.

You can see that you have a mix of trigonometric functions on x and multiples of x. You also have a mix of powers of trigonometric functions.

Let's try applying the power reduction formula Wikipedia gives us for this case: (sin^2X)(cos^2X)=(1-(cos4X))/8


8(sin^2X)(cos^2X) = 1-cos4X
8(1-cos4X)/8=1-cos4X
1-cos4X=1-cos4X

Oops, that was too easy. We should go back and derive that formula to add some more steps



You need to start from the double angle formula. Usually these are accepted as basic formulas. You can find the proof for them online if you need it.

sin2X=2sinXcosX
cos2X=cos^2X-sin^2X

This is the basic double angle formula. You can apply the Pythagorean identity to expand upon it.

The Pythagorean identity is sin^2X+cos^2X=1

substituting for sin^2X:
cos2X=2cos^2X-1

or substituting for cos^2X:
cos2X=1-2sin^2X

Now just isolate sin^2X and cos^2X from those last 2 formulas

sin^2X=(1-cos2X)/2
cos^2X=(cos2X+1)/2

multiply the two equations together

(sin^2X)(cos^2X)=(cos2X-cos^2(2X)+1-cos2X)/4
(sin^2X)(cos^2X)=(-cos^2(2X)+1)/4

Apply the cosine double angle formula again for cos^2X

(sin^2X)(cos^2X)=(1-(cos4X+1)/2)/4
(sin^2X)(cos^2X)=(1-(cos4X+1)/2)/4
(sin^2X)(cos^2X)=(1-(cos4X))/8

And that is called the power reduction formula according to Wikipedia. Just slap it in there and problem solved.

 

[Moto]

Thanks for your help, Dan. I'm always confused when it's written out in text form, but I'll try to figure it out.

EDIT: I think I got it. I'm going to post a scan of my work and maybe you could check it for me Dan? It's a little convoluted, but I think it works.

Give me a second to post it.

 

[Moto]

Okay, I'm not positive that the math gods would approve of this, but here it goes.

Basically what I did was rewrite the problem to isolate cos4x on one side. I then did a u-substitution for cos4x so that it would look more like cos2x because cos2x is a double angle identity. Then I substituted back in for u and tacked on the 2x at the end to balance it out.

Then I split ((sin^2)2x) into two separate sines. I noticed that I had two double angle identities for sine (sin2x=2sinxcosx), so I ran them through the formula. Then I multiplied the two together and then multiplied the whole thing by that negative two, leaving me with what I had on the left hand side after I rewrote the problem.

After all that, I went back to the original identity that I was given in the book and substituted ((1-2sin^2)2x) in for cos4x. I ran it through exactly as I did when I rewrote the problem and ultimately got the same value on both sides.

Can anyone check it out and make sure it holds water?

 

[Dan]

Looks good. I don't know why you showed everything twice though.

 

[Moto]

Looks good. I don't know why you showed everything twice though.

Because I wanted to run it through the original identity just to check it. It's mainly just for me, but I figured I'd explain myself. I take an odd approach sometimes.

Anyways, thanks Dan.

 

[L3N!N]

Americans saying 'Math' really does piss me off.

You're just failing your Englishes course.

 

[Xevrex]

Americans saying 'Math' really does piss me off.

You'd rather us say 'Arithmetic'?

 

[Xendance]

You'd rather us say 'Arithmetic'?

No, I'd rather 'Maths'.