proving a trig

[oded244]

Hey all, this is my first time here The forum looks cool and helpful. To my question:

How can I prove this:

cos^4α - sin^4α = cos2α

Thank you!

 

[galactus]

It's called proving a trig identity.

\(\displaystyle \L\\cos^{4}(x)-sin^{4}(x)=cos(2x)\)

You can use the identity: \(\displaystyle \L\\cos^{2}(x)=\frac{1+cos(2x)}{2}, \;\ sin^{2}(x)=\frac{1-cos(2x)}{2}\)

Then you have:

\(\displaystyle \L\\\left(\frac{1+cos(2x)}{2}\right)^{2}-\left(\frac{1-cos(2x)}{2}\right)^{2}\)

=\(\displaystyle \L\\\frac{1+2cos(2x)+cos^{2}(2x)-1+2cos(2x)-cos^{2}(2x)}{4}\)

=\(\displaystyle \L\\\frac{1}{2}cos(2x)+\frac{1}{2}cos(2x)=cos(2x)\)

Expand and simplify. It should fall into place.

 

[pka]

Here is another way.
\(\displaystyle \L \cos ^4 (x) - \sin ^4 (x) = \left[ {\underbrace {\cos ^2 (x) - \sin ^2 (x)}_{\cos (2x)}} \right]\left[ {\underbrace {\cos ^2 (x) + \sin ^2 (x)}_1} \right] = \cos (2x)\)

 

[galactus]

Yes, the difference of two squares. Nice.

 

[oded244]

i don't think we learned those identities. i understood galactus way but not pkas.

Here's another one that took a couple of hours from my life (im pretty sure that there's a typo in the book but i'll ask you guys to be 100% sure)

 

[galactus]

pka's is easier than mine. He just used the difference of two squares from early algebra.

\(\displaystyle \L\\x^{2}-y^{2}=(x-y)(x+y)\)

Just let \(\displaystyle x=cos^{2}(x), \;\ y=sin^{2}(x)\)

 

[oded244]

the early algebra i understood, but way he divided the left side in cos(2r) and the right in 1 ?

ohh, now i get it. its not division

 

[pka]

i understood galactus way but not pkas.

You do not understand how to factor the difference of two squares?
Are you sure that you ought to be trying mathematics at this level?

 

[oded244]

the early algebra i understood, but way he divided the left side in cos(2r) and the right in 1 ?

ohh, now i get it. its not division

i thought that you used division. i understood. can you check if you can solve the seconded identity ?

 

[ilaggoodly]

the second one is a false identity, i suppose you mean cos^2...it involves the same identity as the former one if thats the case

 

[Deleted member 4993]

i don't think we learned those identities. i understood galactus way but not pkas.

Here's another one that took a couple of hours from my life (im pretty sure that there's a typo in the book but i'll ask you guys to be 100% sure)

Yes - there is a typo.

It should be:

\(\displaystyle cos^2(3x) - sin^2(3x) = cos(6x)\)

The exponent of Cosine has to be changed to 2 (from 3).