Доказать тождество: sin(п/6 + альфа) + sin(п/6- альфа)=cos альфа вычислить: 2ctg30градусов - 2in45градусов + 3sin90градусов -2ctg п/3-3(sing/6)в квадрате +1/2cos п/2

Лебедев Дмитрий Сергеевич

Учитель математики

Проверено учителем

Для доказательства тождества воспользуемся формулой суммы синусов sinx+siny=2sinx+y2cosxy2sine x plus sine y equals 2 sine the fraction with numerator x plus y and denominator 2 end-fraction cosine the fraction with numerator x minus y and denominator 2 end-fraction , что дает нам cosαcosine alpha. Значение выражения составляет 4332+2,25the fraction with numerator 4 the square root of 3 end-root and denominator 3 end-fraction minus the square root of 2 end-root plus 2 comma 25 . ️ Шаг 1: Доказательство тождества Для доказательства тождества sin(π6+α)+sin(π6α)=cosαsine open paren the fraction with numerator pi and denominator 6 end-fraction plus alpha close paren plus sine open paren the fraction with numerator pi and denominator 6 end-fraction minus alpha close paren equals cosine alpha применим формулу суммы синусов: sinx+siny=2sinx+y2cosxy2sine x plus sine y equals 2 sine the fraction with numerator x plus y and denominator 2 end-fraction cosine the fraction with numerator x minus y and denominator 2 end-fraction Пусть x=π6+αx equals the fraction with numerator pi and denominator 6 end-fraction plus alpha и y=π6αy equals the fraction with numerator pi and denominator 6 end-fraction minus alpha . Тогда:

  1. Вычислим полусумму аргументов: x+y2=(π6+α)+(π6α)2=2π62=π6the fraction with numerator x plus y and denominator 2 end-fraction equals the fraction with numerator open paren the fraction with numerator pi and denominator 6 end-fraction plus alpha close paren plus open paren the fraction with numerator pi and denominator 6 end-fraction minus alpha close paren and denominator 2 end-fraction equals the fraction with numerator the fraction with numerator 2 pi and denominator 6 end-fraction and denominator 2 end-fraction equals the fraction with numerator pi and denominator 6 end-fraction Вычислим полуразность аргументов: xy2=(π6+α)(π6α)2=2α2=αthe fraction with numerator x minus y and denominator 2 end-fraction equals the fraction with numerator open paren the fraction with numerator pi and denominator 6 end-fraction plus alpha close paren minus open paren the fraction with numerator pi and denominator 6 end-fraction minus alpha close paren and denominator 2 end-fraction equals the fraction with numerator 2 alpha and denominator 2 end-fraction equals alpha
    Подставим полученные значения в формулу:
    2sinπ6cosα=212cosα=cosα2 sine the fraction with numerator pi and denominator 6 end-fraction cosine alpha equals 2 center dot one-half center dot cosine alpha equals cosine alpha Тождество доказано.

️ Шаг 2: Вычисление значения выражения Подставим табличные значения тригонометрических функций в выражение: 2ctg302sin45+3sin902ctgπ33(sinπ6)2+12cosπ22 ctg 30 raised to the composed with power minus 2 sine 45 raised to the composed with power plus 3 sine 90 raised to the composed with power minus 2 ctg the fraction with numerator pi and denominator 3 end-fraction minus 3 open paren sine the fraction with numerator pi and denominator 6 end-fraction close paren squared plus one-half cosine the fraction with numerator pi and denominator 2 end-fraction

  1. Находим значения функций:
    ctg30=3ctg 30 raised to the composed with power equals the square root of 3 end-root
    sin45=22sine 45 raised to the composed with power equals the fraction with numerator the square root of 2 end-root and denominator 2 end-fraction
    sin90=1sine 90 raised to the composed with power equals 1
    ctgπ3=33ctg the fraction with numerator pi and denominator 3 end-fraction equals the fraction with numerator the square root of 3 end-root and denominator 3 end-fraction
    sinπ6=12sine the fraction with numerator pi and denominator 6 end-fraction equals one-half
    cosπ2=0cosine the fraction with numerator pi and denominator 2 end-fraction equals 0 Подставляем и упрощаем по частям:
    23=232 center dot the square root of 3 end-root equals 2 the square root of 3 end-root
    -222=2negative 2 center dot the fraction with numerator the square root of 2 end-root and denominator 2 end-fraction equals negative the square root of 2 end-root
    31=33 center dot 1 equals 3
    -233=233negative 2 center dot the fraction with numerator the square root of 3 end-root and denominator 3 end-fraction equals negative the fraction with numerator 2 the square root of 3 end-root and denominator 3 end-fraction
    -3(12)2=-314=-0,75negative 3 center dot open paren one-half close paren squared equals negative 3 center dot one-fourth equals negative 0 comma 75
    120=0one-half center dot 0 equals 0 Складываем результаты:
    232+32330,75=(223)32+(30,75)=4332+2,252 the square root of 3 end-root minus the square root of 2 end-root plus 3 minus the fraction with numerator 2 the square root of 3 end-root and denominator 3 end-fraction minus 0 comma 75 equals open paren 2 minus two-thirds close paren the square root of 3 end-root minus the square root of 2 end-root plus open paren 3 minus 0 comma 75 close paren equals the fraction with numerator 4 the square root of 3 end-root and denominator 3 end-fraction minus the square root of 2 end-root plus 2 comma 25

Ответ: Тождество доказано; значение выражения: \frac{4\sqrt{3}}{3} - \sqrt{2} + 2,25. Сообщите, требуется ли вам численное значение данного выражения с плавающей запятой или помощь с другими тригонометрическими формулами.

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