Find sin2x, cos2x, and tan2x if sinx=1/sqrt5 and x terminates in quadrant I

Answer 1

If sin(x) = 1/√5 (opp. side over hypotenuse) , then the adjacent side length is:

adj. side = √((√5)²-1²)

adj. side = 2

cos(x) = 2/√5 (adj. side over hypotenuse)

sin(2x) = 2sin(x)cos(x) = 2(2/√5)(1/√5) = 4/5

cos(2x) = cos²(x) - sin²(x) = (2/√5)² - (1/√5)² = 3/5

tan(2x) = sin(2x)/cos(2x) = (4/5)/(3/5) = 4/3

Answer 2

sinx =1/sqr5 = sqr5/5 = opposite side/hypotenuse, adjacent side = sqr(25-5) = sqr20 = 2sqr5

cosx = 2sqr5/5

sin2x = 2sinxcosx = 2(sqr5/5)(2sqr5/5) = 20/25 = 4/5 = .8

cos2x = cos^2(x)-sin^2(x) = (2sqr5/5)^2 - (sqr5/5)^2 = 20/25-5/25 = 15/25 =3/5 = .6

tan2x = sin2x/cos2x = .8/.6 = 4/3

check the answers

sinx =sqr5/5 = .45

x = sin^-1(.45) = about 26.6 degrees

use a calculator with an inverse trig function

2x = about 53.13

sin53.13 = .8 = 4/5

cos53.13 = .6 = 3/5

tan53.13 = 4/3