Question #37bc4

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Question #37bc4

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Answer 1 · 2017-04-06T23:21:30

$\frac{{(2 {cos}^{2} θ - 1)}^{2}}{{cos}^{4} θ - {sin}^{4} θ} = 1 - 2 {sin}^{2} θ$ $(2cos^2theta-1)^2/(cos^4theta-sin^4theta)=1-2sin^2theta$
$\frac{{cos}^{2} (2 θ)}{({cos}^{2} θ + {sin}^{2} θ) ({cos}^{2} θ - {sin}^{2} θ)} = 1 - 2 {sin}^{2} θ$ $cos^2(2theta)/((cos^2theta+sin^2theta)(cos^2theta-sin^2theta))=1-2sin^2theta$
$\frac{{cos}^{2} (2 θ)}{cos (2 θ)} = 1 - 2 {sin}^{2} θ$ $cos^2(2theta)/cos(2theta)=1-2sin^2theta$
$cos (2 θ) = 1 - 2 {sin}^{2} θ$ $cos(2theta)=1-2sin^2theta$
$1 - 2 {sin}^{2} θ = 1 - 2 {sin}^{2} θ$ $1-2sin^2theta=1-2sin^2theta$
Q.E.D.

Explanation:

The original statement, with $θ$ $theta$ added for sensibility.
$\frac{{(2 {cos}^{2} θ - 1)}^{2}}{{cos}^{4} θ - {sin}^{4} θ} = 1 - 2 {sin}^{2} θ$ $(2cos^2theta-1)^2/(cos^4theta-sin^4theta)=1-2sin^2theta$

Let's work the left hand side.

Using one form of the cosine double angle formula, $cos (2 θ) = 2 {cos}^{2} θ - 1$ $cos(2theta)=2cos^2theta-1$ , the left side becomes
$\frac{{(cos (2 θ))}^{2}}{{cos}^{4} θ - {sin}^{4} θ}$ $(cos(2theta))^2/(cos^4theta-sin^4theta)$ , or equivalently $\frac{{cos}^{2} (2 θ)}{{cos}^{4} θ - {sin}^{4} θ}$ $cos^2(2theta)/(cos^4theta-sin^4theta)$

Then, notice that the denominator is a difference of squares, it can be factored as follows: $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ . In this case $a = {cos}^{2} θ$ $a=cos^2theta$ and $b = {sin}^{2} θ$ $b=sin^2theta$ . So the left side is now
$\frac{{cos}^{2} (2 θ)}{({cos}^{2} θ + {sin}^{2} θ) ({cos}^{2} θ - {sin}^{2} θ)}$ $cos^2(2theta)/((cos^2theta+sin^2theta)(cos^2theta-sin^2theta))$

Now we can use the Pythagorean Identity, ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$ to get
$\frac{{cos}^{2} (2 θ)}{(1) ({cos}^{2} θ - {sin}^{2} θ)}$ $cos^2(2theta)/((1)(cos^2theta-sin^2theta))$ , or equivalently $\frac{{cos}^{2} (2 θ)}{{cos}^{2} θ - {sin}^{2} θ}$ $cos^2(2theta)/(cos^2theta-sin^2theta)$

And if we can use another form of the cosine double angle formula, $cos (2 θ) = {cos}^{2} θ - {sin}^{2} θ$ $cos(2theta)=cos^2theta-sin^2theta$ , we can see that
$\frac{{cos}^{2} (2 θ)}{cos (2 θ)}$ $cos^2(2theta)/cos(2theta)$

Simplifying this gives a friendly
$cos (2 θ)$ $cos(2theta)$

And using the third and final form of the cosine double angle formula, $cos (2 θ) = 1 - 2 {sin}^{2} θ$ $cos(2theta)=1-2sin^2theta$ , the left hand side finally becomes
$1 - 2 {sin}^{2} θ$ $1-2sin^2theta$

And since $1 - 2 {sin}^{2} θ = 1 - 2 {sin}^{2} θ$ $1-2sin^2theta=1-2sin^2theta$ , the left side is equal to the right side and the identity is proven. Q.E.D.

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Question #37bc4

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