1) cos(pi+a)=cospi*cosa-sinpi*sina=-cosa-0=-cosa ч.т.д
2) sin(pi+a)=sinpi*cosa+cospi*sina=0+(-sina)=-sina ч.т.д.
б) sina*sinb+cos(a+b)=sina*sinb+cosa*cosb-sina*sinb=cosa*cosb
г) cosa*cosb+sin(a-b)=cosa*cosb+sina*cosb-cosa*sinb
б) cos 50° cos 5°+sin 50° sin 5° = cos(50-5)=cos45=sqrt(2)/2
г) cos 25° cos 65°-sin 25° sin 65°=cos(25+65)=cos90=0
√6 - 3 → иррациональное<span>
√3 * √5 = </span>√15 → иррациональное<span>
(√5)</span>² = 5 → рациональное<span>
(√6 - 3)</span>² = 6-6√6+9=15-6√6 → иррациональное
Ответ: (√5)² = 5 → рациональное
sin105°*cos75° = sin(180° -75°)*cos75° = sin75°*cos75° =(sin2*75<span>°)/2 =
</span>(sin150°)/2 =(sin(180°- 30°))/2 = (sin30°)<span>/2 =(1/2) /2 =1/4.
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1) sin105°*sin75° = sin(180° -75°)*sin75° = sin75°*sin75° =sin²75°=
(1 -cos2*75°)/2 =(1 -cos150°)/2 = (1 -cos(180° -30°) )/2 = (1+cos30<span>°) /2 =
</span>(2+√3) / 4 .
* * * sin²75° =(sin45°cos30° + cos45°sin30°) ² = ( (1/√2)*(√3)/2 +(1/√2)*(1)/2) ) ² =(1/8) *(√3 +1) ² =(1/8) *(3 +2√3 +1)= (1/4) *(2 +√3 )= (2 +√3 <span>) /4.
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2) </span>4sin(π/6 -β)cos(π/6+β)= 4 *(sin(π/6 -β+π/6+β) + sin(π/6 -β-π/6-β)<span> )/2 =
</span>2 *(sin π/3 + sin( -2β) ) = 2 *( (√3)/2 - sin2β ) =√3 -2 sin<span>2β.
</span>* * * А если преобразование начнем <span>с правой стороны равенства , то</span><span>
3 - 4cos</span>²β = 4(1 - cos²β) -1 =4sin²β -1 =2*2sin²β -1 =2(1 -cos2<span>β) -1 =
</span>2(1 - cos2<span>β -1/2) = </span>2(1/2 -cos2β) = 2(cosπ/3 -cos2β) = 2(cosπ/3 -cos2<span>β) =
</span>- 4sin(π/6- β)*sin(π/6+ β) .
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