ОЧЕНЬ СРОЧНО ПОМОГИТЕ РЕШИТЬ В ВАРИАНТЕ ПО АЛГЕБРЕ НОМЕРА С 1-3!!!!. Created by kristinka75. algebra-ru

ОЧЕНЬ СРОЧНО ПОМОГИТЕ РЕШИТЬ В ВАРИАНТЕ ПО АЛГЕБРЕ НОМЕРА С 1-3!!!!...

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task/29450914 ---------------------

1. Упростить выражение (cosβ +sinβ)/(cosβ-sinβ) - tg( /4 +β)

(cosβ +sinβ) /(cosβ- sinβ) - tg(π/4 +β) =(1 +tqβ) /(1- tgβ) - tg(π/4 +β) =

( tgπ/4 +tqβ) /( tgπ/4 -tgπ/4 * tgβ ) - tg(π/4+β) = tg(π/4+β) - tg(π/4 +β) =0 .

2. Вычислить значение выражение ( 1 - tg²15° )/√3tg15°

Решение ( 1-sin²15°/cos²15°)/√3tg15°=(cos²15 -sin²15)/√3tg15°*cos²15°=

(cos30°/√3sin15°*cos15°=(√3/2 )/√3sin15°*cos15° =1/(2sin15°*cos15°)=1/sin30°=2.

3. Найти Значение выражения: 9(tg(arcsin3/5) )⁻ ²

9(tg(arcsin3/5) )⁻² =9(tg(arctg3/4) )⁻² =9(3/4)⁻²= 9(4/3)²= 9*16/9 = 16.

P.S. α=arcsin3/5 ⇒sinα =3/5;cosα= + √(1-sin²α) =4/5; tgα=3/4 !!! -π/2 ≤ arcsina ≤ π/2

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oganesbagoyan

Verified answer

$\frac{cos\beta + sin\beta}{cos\beta - sin\beta} = \frac{\frac{\sqrt{2}}{2}(cos\beta + sin\beta)}{\frac{\sqrt{2}}{2}(cos\beta - sin\beta)} = \frac{sin\frac{\pi}{4}cos\beta + cos\frac{\pi}{4}sin\beta}{cos\frac{\pi}{4}cos\beta - sin\frac{\pi}{4}sin\beta} = \frac{sin(\frac{\pi}{4} + \beta)}{cos(\frac{\pi}{4} + \beta)} = tg(\frac{\pi}{4} + \beta)$

$\frac{cos\beta + sin\beta}{cos\beta - sin\beta} - tg(\frac{\pi}{4} + \beta) = 0$

$x = 15^0, \frac{1 - tg^2x}{\sqrt{3}tgx} = \frac{1 - \frac{sin^2x}{cos^2x}}{\sqrt{3}\frac{sinx}{cosx}} = \frac{1}{\sqrt{3}} * \frac{cos^2x-sin^2x}{cos^2x} * \frac{cosx}{sinx} = \frac{2}{\sqrt{3}} * \frac{cos^2x-sin^2x}{2sinx*cosx} = \frac{2}{\sqrt{3}} * \frac{cos2x}{sin2x} = \frac{2}{\sqrt{3}} * \frac{cos30^0}{sin30^0} = \frac{2}{\sqrt{3}} * \frac{\sqrt{3}/2}{1/2} = 2$

3 (см. изображение - подобие треугольников)

tg arcsin x = x * 1/√(1-x^2)

x = 3/5 -> tg arcsin 3/5 = 3/5 * 1/√(1 - 9/25) = 3/5 * 1/√(16/25) = 3/5 * 5/4 = 3/4

9 * (tg arcsin 3/5)^(-2) = 9 * (3/4)^(-2) = 9 * 16/9 = 16

1 votes Thanks 1

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