Здравствуйте, помогите пожалуйста . Created by aleksoref. algebra-ru

Здравствуйте, помогите пожалуйста ...

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$1)\ \ cos6a\cdot cos2a-sin6a\cdot sin2a=cos(6a+2a)=cos8a\\\\\\2)\ \ sin(-15^\circ )\cdot cos75^\circ +cos15^\circ \cdot sin75^\circ =sin75^\circ \cdot cos15^\circ -cos75^\circ \cdot sin15^\circ =\\\\=sin(75^\circ -15^\circ )=sin60^\circ =\dfrac{\sqrt3}{2}$

$3)\ \ \dfrac{cos64^\circ \cdot cos4^\circ +sin64^\circ \cdot sin4^\circ }{sin19^\circ \cdot cos41^\circ +sin41^\circ \cdot cos19^\circ }=\dfrac{cos(64^\circ -4^\circ )}{sin(19^\circ +41^\circ )}=\dfrac{cos60^\circ }{sin60^\circ }=tg60^\circ =\sqrt3\\\\\\4)\ \ cos(a-\beta )-2\, sina\cdot sin\beta =cosa\cdot cos\beta +sina\cdot sin\beta-2\, sina\cdot sin\beta =\\\\=cosa\cdot cos\beta -sina\cdot sin\beta =cos(a+\beta )$

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NNNLLL54

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Ответ:

1) cos6a cos2a - sin6a sin2a = cos(6a+2a) = cos8a

2)sin(-15)cos(75) + cos(15)(sin75) = sin(75 + (-15)) = sin60 =

4) cos(a-b) -2sinasinb = cosacosb +sinasinb - 2sinasinb = cosacosb - sinasinb=cos(a+b)

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aleksoref спасибо

aleksoref помогите еще с одним пожалуйста

aleksoref https://znanija.com/task/40986768

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