Question

please help! establish the identity cos^4theta-sin^4theta= cos2theta

Answer 1

Using Euler's formula you can proove any trig identity almost without thinking.
The formula is:\[e^{i \theta}=\cos \theta+isen \theta\]

Answer 2

Thanks! Ill give it a try

Answer 3

Just notice that i is imeginary so we are dealing with complex numbers here.

Answer 4

ok can you show me how to use the formula. I'm not familiar with it.

Answer 5

Ok, sure, to begin with, with complex numbers there are two parts of the number, the real, and the imaginary, if a number is equal to the other, those two parts must be equal as well.
The real part is the part that is alone, and the imaginary part is the part that has the i.
This i is the sqrt(-1) thats why it is called imaginary, because we don't have a clear view of what it is as we have with the other numbers.
Now, thats what you need to know to use the formula.
Now, to apply this to your problem, just clarify if I'm using the right identity.
\[\cos^4 \theta-\sin^4 \theta=\cos(2\theta)\]Is that what you meant?

Answer 6

yes thats right

Answer 7

Ok, now lets apply the formula.
The left part has nothing we can put on the formula, but the right part has.
Our theta in the formula is, in this case 2theta, so we have:
\[e^{i2\theta}\]Now, to apply the formula, you usually write this in two forms and compare them.
\[e^{i2\theta}=e^{i \theta}e^{i \theta}\]Now, what is left is to turn those into sin and cos form by looking at the formula, and comparing them.
\[e^{i2\theta}=\cos(2\theta)+isin(2\theta)\]\[e^{i \theta}e^{i \theta}=(\cos \theta+isin \theta)(\cos \theta+isin \theta)=\cos^2\theta-\sin^2\theta+i2\cos \theta \sin \theta\]That minus appears because i is the square root of minus 1. Now we only need to compare the real and imaginary part because they must be equal. From this, we get two identities.
\[\cos(2\theta)=\cos^2\theta-\sin^2\theta\]and\[\sin(2\theta)=2\sin \theta \cos \theta\]Probably you misread the identity in the first place, because they are ^2 not ^4.
You can do that with any trig identity just as easy. When you see it the first time it gets a lot easier on the next.

Answer 8

OH, you didnt misread.

Answer 9

When we have this identity we only need to use the fact that (a-b)(a+b)=a^2-b^2.
We multiply both sides by cos^2 + sin ^2 wich is 1 and do not affect the cos 2theta so we get:\[\cos(2\theta)=\cos(2\theta)(\cos^2\theta+\sin^2\theta)=(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)=\cos^4\theta-\sin^4\theta\]

Answer 10

Sorry, it didnt fit, but the end was \[\cos^4\theta-\sin^4\theta\]

Answer 11

Did you understand everything? If not ask me because this thing is REALLY helpfull.

Answer 12

thanks. Ive gotta go. you were really helpful