# Thread: Can someone help me prove these two identities?

1. ## Can someone help me prove these two identities?

Can someone help me prove these two identities?

I have to prove these two identities and they're really giving me a hard time :/
I need to see all the steps taken.

1) sin^4 β - cos^4 β = 1 - 2cos^2 β

2) cos Φ/(1 + sinΦ) + (1 + sin Φ)/cos Φ = 2sec Φ

thank youuuu!

2. Originally Posted by dlngoal
Can someone help me prove these two identities?

I have to prove these two identities and they're really giving me a hard time :/
I need to see all the steps taken.

1) sin^4 β - cos^4 β = 1 - 2cos^2 β

2) cos Φ/(1 + sinΦ) + (1 + sin Φ)/cos Φ = 2sec Φ

thank youuuu!
For number 1 use the difference of two squares

$\displaystyle (\sin^2 \beta - \cos^2 \beta)(\sin^2 \beta + \cos^2 \beta)$

Use the Pythagorean identity to solve

3. Hello, dlngoal!

$\displaystyle 2)\;\;\frac{\cos\phi}{1 + \sin\phi} + \frac{1 + \sin\phi}{\cos\phi} \:=\: 2\sec\phi$

On the left side, multiply the first fraction by $\displaystyle \frac{1-\sin\phi}{1-\sin\phi}$

. . $\displaystyle \frac{\cos\phi}{1+\sin\phi}\cdot{\color{blue}\frac {1-\sin\phi}{1-\sin\phi}} + \frac{1+\sin\phi}{\cos\phi} \;\;=\;\;\frac{\cos\phi(1-\sin\phi)}{1-\sin^2\!\phi} + \frac{1+\sin\phi}{\cos\phi}$

. . . . $\displaystyle =\;\;\frac{\cos\phi(1-\sin\phi)}{\cos^2\!\phi} + \frac{1+\sin\phi}{\cos\phi} \;\;=\;\;\frac{1-\sin\phi}{\cos\phi} + \frac{1+\sin\phi}{\cos\phi}$

. . . . . $\displaystyle =\;\;\frac{1-\sin\phi + 1 + \sin\phi}{\cos\phi} \;\;=\;\;\frac{2}{\cos\phi} \;\;=\;\;2\sec\phi$

4. Thank you Soroban!

And e^(i*pi), I was told I had to use trigonometric identity formulas and i have to prove that, for number one,

sin^4 β - cos^4 β = 1 - 2cos^2 β

that the Left side and the right side are truly equal to each other. I still don't understand. :|

5. Originally Posted by dlngoal
Thank you Soroban!

And e^(i*pi), I was told I had to use trigonometric identity formulas and i have to prove that, for number one,

sin^4 β - cos^4 β = 1 - 2cos^2 β

that the Left side and the right side are truly equal to each other. I still don't understand. :|
The identity to use is $\displaystyle \sin^2 A + \cos^2 A = 1$

Using the difference of two squares:

$\displaystyle (\sin ^2 \beta - \cos^2 \beta)(\sin^2 \beta + \cos ^2 \beta) = \sin ^2 \beta - \cos ^2 \beta = (1-\cos^2 \beta) - \cos^2 \beta = 1-2\cos^2 \beta$

6. Originally Posted by dlngoal
Thank you Soroban!

And e^(i*pi), I was told I had to use trigonometric identity formulas and i have to prove that, for number one,

sin^4 β - cos^4 β = 1 - 2cos^2 β

that the Left side and the right side are truly equal to each other. I still don't understand. :|
Hi dlngoal:

I did this for you a few days ago. here's the link:

http://www.mathhelpforum.com/math-he...-identity.html