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Cos(A+B)Cos(A-B)

Prove that: ...

Question

Prove that:
${cos}^{2} α + {cos}^{2} (α + β) - 2 cos α \cdot cos β cos (α + β) = {sin}^{2} β$

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Solution

given, $c o s^{2} α + c o s^{2} (α + β) - 2 c o s α c o s β c o s (α + β)$

$= c o s^{2} α + c o s (α + β) (c o s (α + β) - 2 c o s α c o s β)$

$= c o s^{2} α - c o s (α + β) (2 c o s α c o s β - c o s α c o s β + s i n α s i n β)$

$= c o s^{2} α - c o s (α + β) (c o s α c o s β + s i n α s i n β)$

$= c o s^{2} α - (c o s α c o s β - s i n α s i n β) (c o s α c o s β + s i n α s i n β)$

$= c o s^{2} α - (c o s^{2} α c o s^{2} β - s i n^{2} α s i n^{2} β)$

$= c o s^{2} α - c o s^{2} α c o s^{2} β + s i n^{2} α s i n^{2} β$

$= c o s^{2} α (1 - c o s^{2} β) + s i n^{2} α s i n^{2} β$

$= c o s^{2} α s i n^{2} β + s i n^{2} α s i n^{2} β$

$= s i n^{2} β (c o s^{2} α + s i n^{2} α)$

$= s i n^{2} β$

Hence Proved

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Cos(A+B)Cos(A-B)

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