Question

помогите с формулами двойного угла. ДАЮ 30 БАЛОВ

Answer 1

sin2α =2sinα*cosα ;
cos2α=cos²α -sin²α   || иначе 2cos²α -1   или  1-2sin²α  ||  . 
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12sin11°*cos11°/sin22° =6*(2sin11°*cos11°/sin22°) =6*sin(2*11°/sin22°
6*sin(22°/sin22°=6.
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24(sin²17° -cos°17°)/cos34° = -12*(2(cos°17°-sin²17°)/cos34°) =
 -12*cos(2*17°)/cos34° = -12cos34°/cos34° = -12.
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√3cos²5π/12 -√3sin²5π/12 =√3(cos²5π/12 -sin5π/12) =√3cos(2*5π/12)=
√3cos(5π/6) =√3cos(π -π/6)  =√3*(-cosπ/6) =√3*(-√3/2)  = -1,5.
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8sin5π/12*cos5π/12 =4*(2sin5π/12*cos5π/12) =4*sin(2*5π/12)=
 4*sin(5π/6) =4*sin((π -π/6) =4*sinπ/6=4*1/2 =2.
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√12*cos²5π/12 -√3 =√(3*2²)*cos²5π/12 -√3 =√3*2cos²5π/12 -√3 =
√3(2cos²5π/12 -1) =√3*cos2*5π/12 =√3*cos5π/6=√3cos(π -π/6)  =√3*(-cosπ/6) =√3*(-√3/2)  = -1,5.
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√3 -√12*sin²5π/12 =√3 -√(3*2²) *sin²5π/12 =√3 -√3 *2 *sin²5π/12  =
√3(1 -2 *sin²5π/12) =√3*cos2*5π/12 =√3*cos5π/6=√3cos(π -π/6)  =√3*(-cosπ/6) =√3*(-√3/2)  = -1,5.
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50*sin19°cos19°/sin38° =25*(2sin19°cos19°/sin38°) =25sin2*19°/sin38° =
=25sin38°/sin38° =25.
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√72cos²15π/8 -√18  =√(18*2²)cos²15π/8 -√18)=√18 (2cos²15π/8 -1) =    3√2*cos2*15π/8 =3√2*cos15π/4 =3√2*cos1(4π-π/4 )=3√2*cosπ/4= 
3√2*1/√2 =3.
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(cos²π/8 -sin²π/8)/(√2/2) =(cos2*π/8)/(cosπ/4) =(cosπ/4) /(π/4) =1.
* * *  √2/2  =cosπ/4 * * * 
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2cos²15°*tq15° =2cos²15°*(sin15°/cos15°)=2cos15°*sin15=sin2*15°= sin30° =1/2.
 (устал я)  удачи  !