.Знайти:
1) (6*sin 180) +(5* cos 0) - (2* sin90 )=
2) sin150+cos120=
2. Спростити: 1+соs2A - sin2A =
tq2A - (sin2A*tg2A) =
1) (6*sin 180) +(5* cos 0) - (2* sin90 )=
2) sin150+cos120=
2. Спростити: 1+соs2A - sin2A =
tq2A - (sin2A*tg2A) =
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Відповідь:
1. 1) (6 * sin180) + (5* cos0) - (2* sin90) = (6 * 0) + (5 * 1) - (2 * 1) = 0 + 5 - 2 = 3
2) sin150 + cos120 = -sin30 + (-cos60) = -[tex]\frac{1}{2}[/tex] + (-[tex]\frac{1}{2}[/tex]) = -1
2. 1) 1 + соs2A - sin2A = 1 - 2sinAcosA
2) tq2A - (sin2A * tg2A) = [tex]\frac{8tan^4A}{(1-tan^2A)^2}[/tex]