How do you find the value for sin 2 θ , cos 2 θ , and tan 2 θ given sin θ = (3/8) and θ is in quadrant I?

Answer 1

Since θ is in the first quadrant, it follows that the right triangle is in the first quadrant. Therefore we have,

sinθ=3/8

cosθ=√55/8

tanθ=3/√55

sin2θ= 2 sinθ cosθ

=2 (3/8) (√55/8)

=3√55/32

cos2θ = cos²θ - sin²θ

=(√55/8)² - (3/8)²

^=23/32

tan2θ = 2 tan θ/ 1- tan²θ

= 2 (3/√55) / 1 - (3/√55)²

= 3√55/23