Do you know how much sin65 ° * cos20 ° - sin65 ° * sin20 ° is equal to

Do you know how much sin65 ° * cos20 ° - sin65 ° * sin20 ° is equal to

sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]
cosα·sinβ=(1/2)[sin(α+β)-sin(α-β)]
(sin85°-sin45°-cos85°+cos45°)/2
=(sin85°-cos85°)/2
=(cos5°-sin5°)/2