0.1*√400 - 0.1=0.1*20 -0,1=2-0,1=1.9
(√2+1)²-2√2=(√2)²-2√2+1²-2√2=2+1=3
⁴√27 *(³√-4)³ *⁴√3 =⁴√(3³)* (-4¹/₃)³ *⁴√3=⁴√3⁴ *(-4)=3*(-4)=-12
∛2 *⁴√(-36)² ∛2 * ⁴√6⁴ ∛2 * 6
------------------ = ----------------------- = ----------- =2
∛54 ∛(2*27) 3*∛2
1) -4 sin²α +5 - 4cos²α = -4(sin²α+cos²α) +5 =-4*1 +5 =1.
sin3,5a*sin2,5α+cosa-cos3,5a*cos2,5a = cosα -(cos3,5a*cos2,5a -sin3,5a*sin2,5α) =cosα -cos(3,5α+2,5α) =cosa -cos6α.
2. 2sin2α + 6cos2α =4sinαcosα+6(1-2sin²α) =4(-0,2)cosα +6(1-2*(-0,2)²) =
= -0,8cosα + 0,552 ; cosα = (-/+)√(1-sin²α) = (-/+)√(1-0,04)= (-/+)√(1-0,04) =
=(-/+)4√0,06 =....
3. p₁=4+3(-1) =1 при cosα = - 1 ; p₂=3+4*1 =7 при cosα=1.
1 ≤cosα ≤1 ⇒ p∈[1;7] семь целых чисел: 1;2;3;4;5;6;7.
1)Для сos(a)=0,8:
sin^2(a) = 1 - cos^2(a) = 1 - 0,64 = 0,36
По условию 0<a<pi/2 => a лежит в первой четверти => sin, cos, tg и ctg - положительные
sin(a)=0,6
tg(a)=sin(a)/cos(a)=0,6/0,8=3/4
ctg(a)=4/3
2) sin(a) = 5/13
cos^2(a) = 1 - sin^2(a) = 1 - 25/169 = 144/169
cos(a) = 12/13
tg(a) = (5/13)*(13/12) = 5/12
ctg(a) = 12/5
3) tg(a) = 2,4
tg^2(a) = 5,76
1 + tg^2(a) = 1/cos^2(a) => 1/cos^2(a) = 6,76
cos^2(a) = 1/6,76 = 100/676
cos(a) = 10/26 = 5/13
sin^2(a) = 1 - 25/169 = 144/169
sin(a) = 12/13
ctg(a) = (5/13)*(13/12) = 5/12
4) ctg(a) = 7/24
tg(a) = 24/7
ctg^2(a) = 49/576
1+ctg^2(a) = 1/sin^2(a) => 1/sin^2(a) = 1 + 49/576 = 625/576
sin^2(a) = 576/625
sin(a) = 24/25
cos^2(a) = 1 - sin^2(a) = 1 - 576/625 = 49/625
cos(a) = 7/25
Cos 2L = cos²L - sin²L
cos²L = 1-sin²L тогда
cos2L=1-sin²L-sin²L=1-2sin²L
если sinL =-0.6 то sin²L =-0.6* (-0.6)=0.36
cos2L=1-2*0.36=1-0.72=0.28
Путем подстановки, а) не принадлежит, все остальные принадлежат