Question

dik cos a =6/10 tentukan a)cos a x cos a -sin a x sin a b)sin a x cos a + cos a x sin a c)cos a x tan a + tan a x sin a d)tan a x tan -sin a x sin a​

Answer 1

Jawaban:

Diketahui cos(a) = 6/10

a) cos(a) * cos(a) - sin(a) * sin(a)

Menggunakan identitas trigonometri cos^2(a) - sin^2(a) = cos(2a)

cos(2a) = cos(a + a) = cos(a) * cos(a) - sin(a) * sin(a)

cos(2a) = (6/10) * (6/10) - sqrt(1 - (6/10)^2) * sqrt(1 - (6/10)^2)

cos(2a) = (36/100) - sqrt(1 - 36/100) * sqrt(1 - 36/100)

cos(2a) = 0.36 - sqrt(1 - 0.36) * sqrt(1 - 0.36)

cos(2a) = 0.36 - sqrt(0.64) * sqrt(0.64)

cos(2a) = 0.36 - 0.8 * 0.8

cos(2a) = 0.36 - 0.64

cos(2a) = -0.28

b) sin(a) * cos(a) + cos(a) * sin(a)

Menggunakan identitas trigonometri sin(2a) = 2 * sin(a) * cos(a)

sin(2a) = 2 * (sqrt(1 - cos^2(a))) * cos(a)

sin(2a) = 2 * (sqrt(1 - (6/10)^2)) * (6/10)

sin(2a) = 2 * (sqrt(1 - 36/100)) * (6/10)

sin(2a) = 2 * (sqrt(0.64)) * (6/10)

sin(2a) = 2 * (0.8) * (6/10)

sin(2a) = 0.96

c) cos(a) * tan(a) + tan(a) * sin(a)

cos(a) * tan(a) + tan(a) * sin(a) = (6/10) * (sqrt(1 - cos^2(a)) / cos(a)) + (sqrt(1 - cos^2(a)) / cos(a)) * sin(a)

cos(a) * tan(a) + tan(a) * sin(a) = (6/10) * (sqrt(1 - (6/10)^2)) / (6/10) + (sqrt(1 - (6/10)^2)) / (6/10) * (sqrt(1 - (6/10)^2))

cos(a) * tan(a) + tan(a) * sin(a) = (6/10) * (sqrt(1 - 36/100)) / (6/10) + (sqrt(1 - 36/100)) / (6/10) * (sqrt(1 - 36/100))

cos(a) * tan(a) + tan(a) * sin(a) = (6/10) * (sqrt(0.64)) / (6/10) + (sqrt(0.64)) / (6/10) * (sqrt(0.64))

cos(a) * tan(a) + tan(a) * sin(a) = 0.8 / 0.8 + 0.8 / 0.8 * 0.8

cos(a) * tan(a) + tan(a) * sin(a) = 1 + 0.8

cos(a) * tan(a)