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Marisabel10061999
@Marisabel10061999
October 2021
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oganesbagoyan
Verified answer
а)
cos²15° -sin²15° =cos2*15° =cos30° =√3/2 ;
б)
2tqπ/12/(1+tq²π/12) =sin2*π/12 =sinπ/6 =1/2 ;
в)
sin22,5°*cos22,5° =1/2*sin2*22,5° =1/2*sin45° =(√2)/4 ;
г)
(1 -tq²π/8)/2*tqπ/8 =1/tq2*π/8 =1/tqπ/4 =1/1 =1.
-------------------------------------------------------------
4.
sin(π/4 -α) -sin(π/4 +α) =√2cos(π/2 +α) .
-------------------------------------------------------------
sin(π/4 -α) -sin(π/4 +α) =2sin(π/4 -α-(π/4 +α))/2 *cos(π/4 -α +π/4 +α)/2 =
2sinπ(-α)*cosπ/4 =-2sinπα*√2/2 =√2(-sinα) =√2cos(π/2+α).
или sin(π/4 -α) -sin(π/4 +α) =sinπ/4*cosα -cosπ/4*sinα - (sinπ/4*cosα +cosπ/4*sinα)=
= -2cosπ/4*sinα = -2*√2/2*sinα =√2(-sinα) =√2cos(π/2+α) .
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Answers & Comments
Verified answer
а) cos²15° -sin²15° =cos2*15° =cos30° =√3/2 ;б) 2tqπ/12/(1+tq²π/12) =sin2*π/12 =sinπ/6 =1/2 ;
в) sin22,5°*cos22,5° =1/2*sin2*22,5° =1/2*sin45° =(√2)/4 ;
г) (1 -tq²π/8)/2*tqπ/8 =1/tq2*π/8 =1/tqπ/4 =1/1 =1.
-------------------------------------------------------------
4. sin(π/4 -α) -sin(π/4 +α) =√2cos(π/2 +α) .
-------------------------------------------------------------
sin(π/4 -α) -sin(π/4 +α) =2sin(π/4 -α-(π/4 +α))/2 *cos(π/4 -α +π/4 +α)/2 =
2sinπ(-α)*cosπ/4 =-2sinπα*√2/2 =√2(-sinα) =√2cos(π/2+α).
или sin(π/4 -α) -sin(π/4 +α) =sinπ/4*cosα -cosπ/4*sinα - (sinπ/4*cosα +cosπ/4*sinα)=
= -2cosπ/4*sinα = -2*√2/2*sinα =√2(-sinα) =√2cos(π/2+α) .