Please help me prove this trig identity (please include justification)

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

appreciated

Printable View

- Jun 10th 2012, 11:31 AMdanny88Math Trig Identity
Please help me prove this trig identity (please include justification)

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

appreciated - Jun 10th 2012, 11:56 AMReckonerRe: Math Trig Identity
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 PMdanny88Re: 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 PMReckonerRe: Math Trig Identity
- Jun 10th 2012, 12:24 PMdanny88Re: 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 PMReckonerRe: Math Trig Identity
Thank you.

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 PMdanny88Re: Math Trig Identity
sin^2 A/ cos^2A =tan^2A?

then -cos^2A/cos^2A= -1 ? - Jun 10th 2012, 12:44 PMReckonerRe: Math Trig Identity
- Jun 10th 2012, 12:48 PMdanny88Re: Math Trig Identity
1= sec^2A-tan^2A

but how does that work out? - Jun 10th 2012, 12:51 PMReckonerRe: Math Trig Identity
- Jun 10th 2012, 12:55 PMdanny88Re: Math Trig Identity
thanks a lot of the help, I really appreciate it.