Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do you prove this? $${(1-2\sin^2A)^2 \over \cos^4A-\sin^4A} = {2\cos^2A - 1}$$

share|improve this question
    
Express everything in terms of $u = \cos A$, and go from there... –  vonbrand May 9 '13 at 18:14
    
I don't.${}{}{}$ –  Michael Greinecker May 9 '13 at 21:12

4 Answers 4

$$ \begin{align} \frac{(1-2\sin^2(A))^2}{\cos^4(A)-\sin^4(A)} &=\frac{(1-2\sin^2(A))^2}{(1-\sin^2(A))^2-\sin^4(A)}\\ &=\frac{(1-2\sin^2(A))^2}{1-2\sin^2(A)}\\ &=1-2\sin^2(A)\\[9pt] &=1-2(1-\cos^2(A))\\[9pt] &=2\cos^2(A)-1 \end{align} $$

share|improve this answer

Recall the following expressions for $\cos(2A)$: $$\color{magenta}{\cos(2A)} = \cos^2(A) - \sin^2(A) = \color{red}{2\cos^2(A)-1} = \color{blue}{1-2\sin^2(A)} = \color{green}{\cos^4(A) - \sin^4(A)}$$ Hence, we have $$\dfrac{\left(\color{blue}{1-2\sin^2(A)}\right)^2}{\color{green}{\cos^4(A) - \sin^4(A)}} = \dfrac{\color{magenta}{\cos^2(2A)}}{\color{magenta}{\cos(2A)}} = \color{magenta}{\cos(2A)} = \color{red}{2\cos^2(A)-1}$$

share|improve this answer

${(1-2\sin^2A)^2 \over \cos^4A-\sin^4A}$

$cos^22A \over \ (cos^2A-\sin^2A)(cos^2A+\sin^2A) $

$cos^22A \over \ (cos^2A-\sin^2A) \ *\ 1 $

$cos^22A \over \ (cos2A) $

cos2A

${2\cos^2A - 1}$

share|improve this answer

Observe that we need to eliminate $\sin A$

So using $\sin^2A=1-\cos^2A,$

$1-2\sin^2A=1-2(1-\cos^2A)=2\cos^2A-1$

and $\cos^4A-\sin^4A=(\cos^2A-\sin^2A)(\cos^2A+\sin^2A)=1\cdot\{\cos^2A-(1-\cos^2A)\}=2\cos^2A-1$

or $\cos^4A-\sin^4A=\cos^4A-(1-\cos^2A)^2=2\cos^2A-1$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.