Cos(α-β)-2sinα sinβ/2sinα cosβ-sin(α-β)

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вопрос

To simplify the expression, we can use the trigonometric identities:

Substitute these identities into the expression:

cos(α-β) - 2sinαsinβ / 2sinαcosβ - sin(α-β)
= (cosαcosβ + sinαsinβ) - 2sinαsinβ / 2sinαcosβ - (sinαcosβ - cosαsinβ)
= cosαcosβ + sinαsinβ - 2sinαsinβ / 2sinαcosβ - sinαcosβ + cosαsinβ
= cosαcosβ + sinαsinβ - 2sinαsinβ / 2sinαcosβ - sinαcosβ + cosαsinβ

Now, simplify each term:
= cosαcosβ + sinαsinβ - 2sinαsinβ / 2sinαcosβ - sinαcosβ + cosαsinβ
= cosαcosβ + sinαsinβ - 2sinαsinβ / 2sinαcosβ - sinαcosβ + cosαsinβ
= cosαcosβ - sinαsinβ / 2sinαcosβ - sinαcosβ + cosαsinβ

Therefore, the simplified expression is:

(cosαcosβ - sinαsinβ) / (2sinαcosβ - sinαcosβ) + cosαsinβ
= cos(α + β) + cosαsinβ

This is the simplified form of the given expression.