1. sin^2 x = 0,9 cos^2 x = 1 - sin^2 x = 1 - 0,9 = 0,1 11 - 99cos^2 x = 11 - 99*0,1 = 11 - 9,9 = 1,1
2. tg a = 2; 1/cos^2 a = 1 + tg^2 a = 1 + 4 = 5 cos^2 a = 1/5; sin^2 a = 1 - cos^2 a = 1 - 1/5 = 4/5 a ∈ [pi; 2pi], поэтому sin a < 0 sin a = -2/√5 sin a / (√5) = -(2/√5) / (√5) = -2/5
3. sin (33pi/4) = sin (32pi/4 + pi/4) = sin (pi/4) = √2/2 cos (34pi/3) = cos (30pi/3 + 4pi/3) = cos (4pi/3) = -cos (pi/3) = -1/2 √32*sin (33pi/4)*cos (34pi/3) = 4√2*√2/2*(-1/2) = -4/2*(√2*√2)/2 = -2
Answers & Comments
Verified answer
1. sin^2 x = 0,9cos^2 x = 1 - sin^2 x = 1 - 0,9 = 0,1
11 - 99cos^2 x = 11 - 99*0,1 = 11 - 9,9 = 1,1
2. tg a = 2; 1/cos^2 a = 1 + tg^2 a = 1 + 4 = 5
cos^2 a = 1/5; sin^2 a = 1 - cos^2 a = 1 - 1/5 = 4/5
a ∈ [pi; 2pi], поэтому sin a < 0
sin a = -2/√5
sin a / (√5) = -(2/√5) / (√5) = -2/5
3. sin (33pi/4) = sin (32pi/4 + pi/4) = sin (pi/4) = √2/2
cos (34pi/3) = cos (30pi/3 + 4pi/3) = cos (4pi/3) = -cos (pi/3) = -1/2
√32*sin (33pi/4)*cos (34pi/3) = 4√2*√2/2*(-1/2) = -4/2*(√2*√2)/2 = -2
4. Формулы синуса и косинуса двойных углов:
sin 2a = 2sin a*cos a; cos 2a = 2cos^2 a - 1
Подставляем в пример
1 - 2cos^2 (111) = -cos (222) = -cos (180 + 42) = cos 42
cos^2 (114) = cos^2 (180 - 66) = (-cos 66)^2 = cos^2 (66)
10cos^2 (66)*tg (66) = 10cos^2 (66)*sin (66)/cos (66) =
= 10cos (66)*sin (66) = 5sin (132) = 5sin (90 + 42) = 5cos 42
Получаем
[1 - 2cos^2 (111)] / [10cos^2 (114)*tg (66)] = cos 42 / (5cos 42) = 1/5
Verified answer
Всё подробно написала в решении.