# Can't figure out this Trig. proof: cos &#x2061;<!-- ⁡ --> ( x + y ) cos

Caleb Proctor 2022-07-04 Answered
Can't figure out this Trig. proof: $\mathrm{cos}\left(x+y\right)\mathrm{cos}\left(x-y\right)={\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(y\right)$
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## Answers (1)

sniokd
Answered 2022-07-05 Author has 22 answers
HINT: Notice,
$LHS=\mathrm{cos}\left(x+y\right)\mathrm{cos}\left(x-y\right)=\left(\mathrm{cos}x\mathrm{cos}y-\mathrm{sin}x\mathrm{sin}y\right)\left(\mathrm{cos}x\mathrm{cos}y+\mathrm{sin}x\mathrm{sin}y\right)$
$=\left(\mathrm{cos}x\mathrm{cos}y{\right)}^{2}-\left(\mathrm{sin}x\mathrm{sin}y{\right)}^{2}$
$={\mathrm{cos}}^{2}x{\mathrm{cos}}^{2}y-{\mathrm{sin}}^{2}x{\mathrm{sin}}^{2}y$
$={\mathrm{cos}}^{2}x\left(1-{\mathrm{sin}}^{2}y\right)-\left(1-{\mathrm{cos}}^{2}x\right){\mathrm{sin}}^{2}y$
$={\mathrm{cos}}^{2}x-{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}y-{\mathrm{sin}}^{2}y+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}y$
$={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}y$

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