# How do I verify the following? cos(x+y)cos(x-y)=cos^2x-sin^2y

Then teach the underlying concepts
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#### Explanation

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#### Explanation:

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61
vince Share
Mar 10, 2015

Hello,

Use usual formulas :

$\cos \left(x + y\right) = \cos \left(x\right) \cos \left(y\right) - \sin \left(x\right) \sin \left(y\right) = A + B$,
$\cos \left(x - y\right) = \cos \left(x\right) \cos \left(y\right) + \sin \left(x\right) \sin \left(y\right) = A - B$.

So, $\cos \left(x + y\right) \cos \left(x - y\right) = \left(A + B\right) \left(A - B\right) = {A}^{2} - {B}^{2}$,

it means

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

But you know that ${\cos}^{2} \left(T\right) + {\sin}^{2} \left(T\right) = 1$, so,

${\cos}^{2} \left(x\right) {\cos}^{2} \left(y\right) - {\sin}^{2} \left(x\right) {\sin}^{2} \left(y\right) = {\cos}^{2} \left(x\right) \left(1 - {\sin}^{2} \left(y\right)\right) - \left(1 - {\cos}^{2} \left(x\right)\right) {\sin}^{2} \left(y\right)$.

Develop and find

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

Simplify :

${\cos}^{2} \left(x\right) {\cos}^{2} \left(y\right) - {\sin}^{2} \left(x\right) {\sin}^{2} \left(y\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(y\right)$.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

18
vince Share
Mar 10, 2015

Hello,

Use usual formulas :

$\cos \left(x + y\right) = \cos \left(x\right) \cos \left(y\right) - \sin \left(x\right) \sin \left(y\right) = A + B$,
$\cos \left(x - y\right) = \cos \left(x\right) \cos \left(y\right) + \sin \left(x\right) \sin \left(y\right) = A - B$.

So, $\cos \left(x + y\right) \cos \left(x - y\right) = \left(A + B\right) \left(A - B\right) = {A}^{2} - {B}^{2}$,

it means

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

But you know that ${\cos}^{2} \left(T\right) + {\sin}^{2} \left(T\right) = 1$, so,

${\cos}^{2} \left(x\right) {\cos}^{2} \left(y\right) - {\sin}^{2} \left(x\right) {\sin}^{2} \left(y\right) = {\cos}^{2} \left(x\right) \left(1 - {\sin}^{2} \left(y\right)\right) - \left(1 - {\cos}^{2} \left(x\right)\right) {\sin}^{2} \left(y\right)$.

Develop and find

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

Simplify :

${\cos}^{2} \left(x\right) {\cos}^{2} \left(y\right) - {\sin}^{2} \left(x\right) {\sin}^{2} \left(y\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(y\right)$.

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