# Trig Identity - Im so confused!

• Oct 14th 2009, 08:54 AM
repete
Trig Identity - Im so confused!
This is my first post. Up until this point I have been able to figure out things by myself but these trig identities are killing me. The ones in my textbook I can figure out but the identities my teacher gives are not as clear to me. Here is the one I am I currently stuck on.

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

I'm not sure where to start. Perhaps the right side?

I hope I formatted it right. I don't know how to make it look pretty.

Thanks for any help!
• Oct 14th 2009, 08:57 AM
Quote:

Originally Posted by repete
This is my first post. Up until this point I have been able to figure out things by myself but these trig identities are killing me. The ones in my textbook I can figure out but the identities my teacher gives are not as clear to me. Here is the one I am I currently stuck on.

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

I'm not sure where to start. Perhaps the right side?

I hope I formatted it right. I don't know how to make it look pretty.

Thanks for any help!

Is this the identity you're trying to prove:

$\displaystyle 1-\frac{1}{2sin(2x)}=\frac{sin^3(x)+cos^3(x)}{sin(x) +cos(x)}$

??
• Oct 14th 2009, 09:02 AM
repete
How do you make it look like that? I'm guessing its code?

That's very close except the left side is...

1-(1/2sin(2x))

Hopefully the parentheses makes it clear.
• Oct 14th 2009, 09:03 AM
The first thing you can do to the right side is cancel out the denominator by factoring the numerator as the sum of two cubes.
• Oct 14th 2009, 09:05 AM
Quote:

Originally Posted by repete
How do you make it look like that? I'm guessing its code?

That's very close except the left side is...

1-(1/2sin(2x))

Hopefully the parentheses makes it clear.

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

There is a PDF file that you can view.
• Oct 14th 2009, 09:13 AM
The right side can be simplified down to this:

$\displaystyle \frac{sin^3(x)+cos^3(x)}{sin(x)+cos(x)}=\frac{[sin(x)+cos(x)][cos^2(x)-sin(x)cos(x)+sin^2(x)]}{sin(x)+cos(x)}$

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

$\displaystyle =1-sin(x)cos(x)$

I guess your left side must be:

$\displaystyle 1-\frac{1}{2}sin(2x)$

$\displaystyle =1-\frac{1}{2}(2sin(x)cos(x))$

$\displaystyle =1-sin(x)cos(x)$
• Oct 14th 2009, 09:17 AM
repete
Thanks adkinsjr! I didn't even think about factoring it. I was able to prove it myself with that step in. I'm sure I will be back with more identities. Thanks again!