Question #12b6c

3 Answers
Sep 22, 2017

See the proof below

Explanation:

We use

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

$L H S = {\left(2 r \sin x \cos x\right)}^{2} + {r}^{2} {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2}$

$= \left(4 {r}^{2} {\sin}^{2} x {\cos}^{2} x\right) + {r}^{2} {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2}$

$= 4 {r}^{2} {\sin}^{2} x {\cos}^{2} x + {r}^{2} \left({\cos}^{4} x + {\sin}^{4} x - 2 {\cos}^{2} x {\sin}^{2} x\right)$

$= {r}^{2} \left({\cos}^{4} x + {\sin}^{4} x + 2 {\sin}^{2} x {\cos}^{2} x\right)$

$= {r}^{2} {\left({\cos}^{2} x + {\sin}^{2} x\right)}^{2}$

$= {r}^{2}$

$= R H S$

$Q E D$

Sep 22, 2017

$L H S = {\left(2 r \sin x \cos x\right)}^{2} + {r}^{2} {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2}$

$= {\left(r \sin 2 x\right)}^{2} + {r}^{2} {\left(\cos 2 x\right)}^{2}$

$= {r}^{2} \left({\sin}^{2} 2 x + {\cos}^{2} 2 x\right)$

$= {r}^{2} = R H S$

Sep 22, 2017

This question has multiple steps, so skip ahead to the explanation.

Explanation:

So, we are trying to prove: ${\left(2 r \sin x \cos x\right)}^{2} + {r}^{2} {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2} = {r}^{2}$

We can ignore the right side of the equal sign. First, let's move the $r$ outside the brackets or radicals.

${r}^{2} {\left(2 \sin x \cos x\right)}^{2} + {r}^{2} {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2}$
${r}^{2} \left({\left(2 \sin x \cos x\right)}^{2} + {\left({\cos}^{2} x - {\sin}^{2} x\right)}^{2}\right)$

Using the proof ${\cos}^{2} x - {\sin}^{2} x = \cos \left(2 x\right)$${.}^{1}$
${r}^{2} \left({\left(2 \sin x \cos x\right)}^{2} + {\left(\cos \left(2 x\right)\right)}^{2}\right)$
${r}^{2} \left({\left(2 \sin x \cos x\right)}^{2} + {\cos}^{2} \left(2 x\right)\right)$

Expand.
${r}^{2} \left(4 {\sin}^{2} x {\cos}^{2} x + {\cos}^{2} \left(2 x\right)\right)$

Use the proof $\cos \left(2 x\right) = 2 {\cos}^{2} x - 1$
${r}^{2} \left(4 {\sin}^{2} x {\cos}^{2} x + {\left(2 {\cos}^{2} x - 1\right)}^{2}\right)$
${r}^{2} \left(4 {\sin}^{2} x {\cos}^{2} x + 4 {\cos}^{4} x - 4 {\cos}^{2} x + 1\right)$

Group $\cos x$
${r}^{2} \left(4 {\cos}^{2} x \left({\sin}^{2} x + {\cos}^{2} x - 1\right) + 1\right)$

Use the proof ${\sin}^{2} x + {\cos}^{2} x = 1$
${r}^{2} \left(4 {\cos}^{2} x \left(1 - 1\right) + 1\right)$
${r}^{2} \left(4 {\cos}^{2} x \left(0\right) + 1\right)$
${r}^{2} \left(1\right)$
${r}^{2}$

Comparing with the right side of the equal sign...
${r}^{2} = {r}^{2}$.

And we're done!

Footnote 1: instead of substituting ${\cos}^{2} x - {\sin}^{2} x = \cos \left(2 x\right)$ into $\cos \left(2 x\right) = 2 {\cos}^{2} x - 1$, you could go from ${\cos}^{2} x - {\sin}^{2} x$ to $2 {\cos}^{2} x - 1$ without making the substitution.
${\cos}^{2} x - {\sin}^{2} x$.

After rearranging the proof: $1 = {\sin}^{2} x + {\cos}^{2} x \to - {\sin}^{2} x = {\cos}^{2} x - 1$ We can plug this in.

${\cos}^{2} x - {\cos}^{2} x - 1 = 2 {\cos}^{2} x - 1$.