Ответ:

Объяснение:

Во-первых, напишем сами формулы двойного аргумента:

sin 2x = 2sin x*cos x

cos 2x = cos^2 x - sin^2 x = 2cos^2 x - 1 = 1 - 2sin^2 x

tg 2x = 2tg x / (1 - tg^2 x)

И не забываем основное тригонометрическое тождество:

sin^2 x + cos^2 x = 1

Теперь решаем примеры.

1) sin x = 1/3, x - из 2 четверти, значит, cos x < 0.

cos^2 x = 1 - sin^2 x = 1 - (1/3)^2 = 1 - 1/9 = 8/9

cos x = -√(8/9) = -2√2/3

sin 2x = 2sin x*cos x = 2*1/3*(-2√2/3) = -4√2/9

2) cos x = -1/3

cos 2x = 2cos^2 x - 1 = 2(-1/3)^2 - 1 = 2*1/9 - 1 = 2/9 - 1 = -7/9

3) tg x = 2

tg 2x = 2tg x / (1 - tg^2 x) = 2*2/(1 - 2^2) = 4/(1 - 4) = -4/3

4) 2sin(π/12)*cos(π/12) = sin(π/6) = 1/2

5) sin 120° = 2sin 60°*cos 60° = 2*√3/2*1/2 = √3/2

6) cos x = 1/6, x - из 1 четверти, значит, sin x > 0

sin^2 x = 1 - cos^2 x = 1 - (1/6)^2 = 1 - 1/36 = 35/36

sin x = √(35/36) = √35/6

sin 2x = 2sin x*cos x = 2*√35/6*1/6 = √35/18

7) sin x = -2/7

cos 2x = 1 - 2sin^2 x = 1 - 2(-2/7)^2 = 1 - 2*4/49 = 1 - 8/49 = 41/49

8) tg x = 3

tg 2x = 2tg x / (1 - tg^2 x) = 2*3/(1 - 3^2) = 6/(1-9) = 6/(-8) = -3/4

9) 2sin 45°*cos 45° = sin 90° = 1

10) cos 120° = 2cos^2 60° - 1 = 2*(1/2)^2 - 1 = 2/4 - 1 = -1/2

11) sin x = -2/5, x - из 3 четверти, значит, cos x < 0

cos^2 x = 1 - sin^2 x = 1 - (-2/5)^2 = 1 - 4/25 = 21/25

cos x = -√(21/25) = -√21/5

sin 2x = 2sin x*cos x = 2(-2/5)(-√21/5) = 4√21/25

12) cos x = 0,3

cos 2x = 2cos^2 x - 1 = 2*0,3^2 - 1 = 2*0,09 - 1 = 0,18 - 1 = -0,82

13) tg x = -2

tg 2x = 2tg x / (1 - tg^2 x) = 2(-2)/(1 - (-2)^2) = -4/(1 - 4) = 4/3