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znanija.com/task/35188417

1.а)       * * * cosα*cosβ = ( cos(α -β) +cos(α+β) ) /2

            * * *  sinα*cosβ  = ( sin(α+β) +sin(α - β)  ) /2

cos18°*cos72° - sin63°cos27°  =  

( cos(72°- 18°) +cos(72°+18°) ) /2  - ( sin(63°+27°) +sin(63° -27°) ) /2 =

(cos54° + cos90° )/2  - ( sin90° +sin36 ) /2 =

( cos(90°- 36°) + 0 )/2 - ( 1 +sin36°)/2  = ( sin36° - (1+sin36°) ) /2  = - 1/2.

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2.  вычислить       cos10°*cos50°*cos70°

можно разными комбинациями

cos10°*cos50°*cos70° =(1/2) ( cos(50°+10°) +cos(50°- 10°) )*cos70°=

(1/2) (cos60° +cos40°)*cos70° =(1/2) ( (1/2)*cos70° ° +cos40°*cos70° )=

=(1/2) ( (1/2)*cos70°  +(1/2)( cos(70°+40°)+cos(70°-40°) ) ) =

(1/4) ( cos70°  +cos110° + cos30° ) =1/4) ( cos70°  +cos(180° -70°) + √3 /2 )

(1/4) ( cos70°- cos70° + √3 /2 )  = √3 /8

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3.   упростите

cos(α+β)*cos(α -β) - cos²α -cos²β =

(1/2) ( cos(α+β+α -β ) +cos(α+β- α +β) ) + - cos²α - cos²β =

(1/2)cos2α+ (1/2)cos2β - cos²α - cos²β =

(1/2)( cos²α - sin²α + cos²β - sin²β  - 2cos²α - 2cos²β ) =

(1/2)( - cos²α - sin²α - sin²β - cos²β  ) =

(-1/2) ( (cos²α + sin²α) + (sin²β+ cos²β) ) = (-1/2) ( 1 + 1 ) = - 1

Второй способ       * * *    cos²γ  = (1+cos2γ )/2 * * *

cos(α+β)*cos(α -β) - cos²α -cos²β =

cos(α+β)*cos(α -β) - (1+cos2α) /2   - (1+cos2β) /2 =

cos(α+β)*cos(α -β) - 1  - (cos2α + cos2β) /2 =

cos(α+β)*cos(α -β) - 1  - (2cos(α + β) *cos(α- β) ) /2 =

cos(α+β)*cos(α -β) -  1  - cos(α + β) *cos(α- β)  =  - 1