помогите с формулами двойного угла. ДАЮ 30 БАЛОВ. Created by darinakononova. algebra-ru

помогите с формулами двойного угла. ДАЮ 30 БАЛОВ...

darinakononova @darinakononova

November 2021 1 16 Report

помогите с формулами двойного угла. ДАЮ 30 БАЛОВ

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oganesbagoyan

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Sin2α =2sinα*cosα ;
cos2α=cos²α -sin²α || иначе 2cos²α -1 или 1-2sin²α  || .
==================
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.
---
√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.
---
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.
---
√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.
---
√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.
---
50*sin19°cos19°/sin38° =25*(2sin19°cos19°/sin38°) =25sin2*19°/sin38° =
=25sin38°/sin38° =25.
---
√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 * * *
---
2cos²15°*tq15° =2cos²15°*(sin15°/cos15°)=2cos15°*sin15=sin2*15°= sin30° =1/2.
(устал я) удачи !

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