# Math Trig Identity

• Jun 10th 2012, 11:31 AM
danny88
Math Trig Identity

(sinA^4 - cosA^4) secA^2 =secA^2-2

appreciated
• Jun 10th 2012, 11:56 AM
Reckoner
Re: Math Trig Identity
Quote:

Originally Posted by danny88
(sinA^4 - cosA^4) secA^2 =secA^2-2

Are you sure that that's not supposed to be $\displaystyle \left(\sin^4A-\cos^4A\right)\sec^2A = \sec^2A-2\mathrm?$

If that is what you are trying to prove, begin by factoring the $\displaystyle \sin^4A-\cos^4A$ as a difference of squares. Then use the Pythagorean identity and the rest should be obvious.
• Jun 10th 2012, 12:18 PM
danny88
Re: Math Trig Identity
I would really appreciate if you would show me step by step because I seem to be making an error in my steps.
• Jun 10th 2012, 12:20 PM
Reckoner
Re: Math Trig Identity
Quote:

Originally Posted by danny88
I would really appreciate if you would show me step by step because I seem to be making an error in my steps.

Well, how about you post the work that you have done so that I can see where you went wrong?
• Jun 10th 2012, 12:24 PM
danny88
Re: Math Trig Identity
(sin^2 A+cos^2 A)(sin^2A-cos^2A) * sec^2A
(1)*(sin^2A-cos^2A) * sec^2A
1*(sinA-cosA)^2 * sec^2A
1*(1-2sinAcosA)* sec^2A
• Jun 10th 2012, 12:35 PM
Reckoner
Re: Math Trig Identity
Thank you.
Quote:

Originally Posted by danny88
1*(sinA-cosA)^2 * sec^2A

This is your error. We cannot distribute powers over addition or subtraction and, likewise, we cannot "factor out" a power like that. $\displaystyle a^2+b^2$ is not necessarily equal to $\displaystyle (a+b)^2$. Instead, convert the secants to cosines:

$\displaystyle \left(\sin^4A - \cos^4A\right)\sec^2A$

$\displaystyle =\left(\sin^2A + \cos^2A\right)\left(\sin^2A - \cos^2A\right)\sec^2A$

$\displaystyle =\left(\sin^2A - \cos^2A\right)\sec^2A$

$\displaystyle =\frac{\sin^2A - \cos^2A}{\cos^2A}$

$\displaystyle =\frac{\sin^2A}{\cos^2A} - \frac{\cos^2A}{\cos^2A}$

Can you continue from here?
• Jun 10th 2012, 12:41 PM
danny88
Re: Math Trig Identity
sin^2 A/ cos^2A =tan^2A?
then -cos^2A/cos^2A= -1 ?
• Jun 10th 2012, 12:44 PM
Reckoner
Re: Math Trig Identity
Quote:

Originally Posted by danny88
sin^2 A/ cos^2A =tan^2A?
then -cos^2A/cos^2A= -1 ?

Correct. So we have $\displaystyle \tan^2A-1.$

Is there an identity we can use that relates $\displaystyle \tan^2A$ with $\displaystyle \sec^2A\,?$
• Jun 10th 2012, 12:48 PM
danny88
Re: Math Trig Identity
1= sec^2A-tan^2A
but how does that work out?
• Jun 10th 2012, 12:51 PM
Reckoner
Re: Math Trig Identity
Quote:

Originally Posted by danny88
1= sec^2A-tan^2A
but how does that work out?

Solve for $\displaystyle \tan^2A:$

$\displaystyle \tan^2A+1=\sec^2A\Rightarrow\tan^2A=\sec^2A-1$

Now substitute that into our expression:

$\displaystyle \tan^2A-1$

$\displaystyle =\left(\sec^2A-1\right)-1$

$\displaystyle =\sec^2A-2.$
• Jun 10th 2012, 12:55 PM
danny88
Re: Math Trig Identity
thanks a lot of the help, I really appreciate it.