1)Вычислите значение выражения tga*cos^2a, если sina = 3/5 и 0° 2)Вычислите значение выражения ctga-sin^2a, если cosa= 5/13 и 0°
Решение 1) sina = 3/5 cosa = (+ -) √(1 - sin²a) = (+ -)√(1 - (3/5)²)) = (+ -)√(16/25) = (+ -) (4/5) tga = sina/cosa tga = 3/5 : 4/5 = 3/4 tga = 3/5 : (-4/5) = - 3/4 tga * cos²a = (3/4) * (4/5)² = (3*16)/(16/25) = 12/25 tga * cos²a = ( - 3/4) * (4/5)² = (- 3*16)/(16/25) = - 12/25 tga * cos²a = tg0 * cos²0 = 0 * 1 = 0 2) cosa = 5/13 sinx = (+ -)√(1 - cos²a) = (+ -)√(1 - (5/13)²) = (+ -) √(144/169) = (+ -) (12/13) ctga = cosa/sina ctga = 5/13 : (12/13) = 5/12 ctga = 5/13 : (- 12/13) = - 5/12 ctga * sin²a = 5/12 * 144/169 = 60/169 ctga * sin²a = - 5/12 * 144/169 = - 60/169
ctga * sin²a = cos0*sin0 = 1*0 = 0