Proving indentities

• Dec 21st 2011, 03:28 PM
Furyan
Proving indentities
Hello

I have been having more then some trouble proving some identities. Mostly after some time I have found that I've been missing something very simple and I'm sure it's the same with these, but I'd appreciate some help.

The questions I'm working on are in a section dealing with only the following four identities:

$\tan{x} = \dfrac {\sin x}{\cos x}$

$\sin^2{x} + \cos^2{x} = 1$

$\1 + \tan^2{x} = \sec^2{x}$

$\1 + \cot^2{x} = cosec^2{x}$

I can't do either of these

$\cos^4{x} - \sin^4{x} = \cos^2{x} - \sin^2{x}$

$\sin{x} + \cos{x} = \dfrac {\1 – 2 \cos^2x}{\sin x -\cos x}$

A hint at how to approach these would be very welcome.

Thank you
• Dec 21st 2011, 03:38 PM
Plato
Re: Proving indentities
Quote:

Originally Posted by Furyan
I can't do either of these
$\cos^4{x} - \sin^4{x} = \cos^2{x} - \sin^2{x}$

$\sin{x} + \cos{x} = \dfrac {1 - 2 \cos^2x}{\sin x -\cos x}$

A hint at how to approach these would be very welcome.

Factor the first FHS: T^4-T^4=(T^2+T^2)(T^2-T^2)

In the second, multiply numerator and denominator by $\sin(x)-\cos(x).$
• Dec 21st 2011, 04:17 PM
Furyan
Re: Proving indentities
Thank you Plato,

The difference of two squares. I should have seen that and will try it, thank you.

For the second, I have been trying to do these just using identities. Do I need, in general, to be thinking about manipulating in the way you have shown? I thought I could only use identities.

Thank you.
• Dec 21st 2011, 04:23 PM
e^(i*pi)
Re: Proving indentities
Quote:

Originally Posted by Furyan
Thank you Plato,

The difference of two squares. I should have seen that and will try it, thank you.

For the second, I have been trying to do these just using identities. Do I need, in general, to be thinking about manipulating in the way you have shown? I thought I could only use identities.

Thank you.

You can always use the difference of two squares once you change the numerator a bit:

$\dfrac{1-2\cos^2(x)}{\sin(x) - \cos(x)} = \dfrac{1-\cos^2(x) - \cos^2(x)}{\sin(x)-\cos(x)}$

Now what can you tell me about the first two terms in the numerator?
• Dec 21st 2011, 04:36 PM
Furyan
Re: Proving indentities
Hello

I can tell you that the first two terms of the numerator on the RHS are equal to $\sin^2{(x)}$
• Dec 21st 2011, 04:44 PM
Furyan
Re: Proving indentities
Now I think I can see what you are hinting at and it's very helpful, thank you, but I will have to look at it in more detail in the morning.

Thank you
• Dec 21st 2011, 05:34 PM
Prove It
Re: Proving indentities
Quote:

Originally Posted by Furyan
$\sin^2{x} - \cos^2{x} = 1$

Surely the identity you were given was \displaystyle \begin{align*} \sin^2{x} + \cos^2{x} \equiv 1 \end{align*}, not what you wrote above...
• Dec 22nd 2011, 05:10 AM
e^(i*pi)
Re: Proving indentities
Quote:

Originally Posted by Furyan
Hello

I can tell you that the first two terms of the numerator on the RHS are equal to $\sin^2{(x)}$

Quote:

Originally Posted by Furyan
Now I think I can see what you are hinting at and it's very helpful, thank you, but I will have to look at it in more detail in the morning.

Thank you

You end up with $\dfrac{\sin^2(x) - \cos^2(x)}{\sin(x)-\cos(x)} = \dfrac{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}{\sin(x)-\cos(x)}$
• Dec 22nd 2011, 05:44 AM
Furyan
Re: Proving indentities
Quote:

Originally Posted by e^(i*pi)
You end up with $\dfrac{\sin^2(x) - \cos^2(x)}{\sin(x)-\cos(x)} = \dfrac{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}{\sin(x)-\cos(x)}$

Thank you. I have been trying to reply for ages, but am having technical problems. Your posts have been very helpful.
• Dec 22nd 2011, 05:48 AM
Furyan
Re: Proving indentities
Thank you all for your posts they have all been really helpful and I'm very grateful. I'm sorry for not replying to all of them individually, but I'm experiencing technical difficulties. I will try using the methods you have shown me when proving the rest of the identities I'm working on.

Thanks again.