6. Given 𝑠𝑖𝑛 𝜃 = − 1/5 and 𝑐𝑜𝑠 𝜃 > 0 find (a) sin 20, and (b) cos 20

Answer 1

θ is in QIV

cos θ = 2√6

sin 2θ = 2 sin θ cos θ

sin 2θ = 2(-1/5)(2√6)

cos 2θ = 1 - sin² θ

cos 2θ = 1 - (-1/5)²

Answer 2

Given sin(𝜃) = −1/5, cos(𝜃) > 0, and sin²(𝜃) + cos²(𝜃) = 1 we have all of the following:

(a) sin(2𝜃) = 2 sin(𝜃) cos(𝜃) = 2 sin(𝜃) √[ 1 - sin²(𝜃) ] = 2 (-1/5) √[ 1 - (-1/5)² ] = (-2/5) √[ 25/25 - 1/25 ]

= (-2/5) √[ 24/25 ] = (-2/5) √(24) / 5 = -2 √(4 * 6) / 25 = -2 √(4) * √(6) / 25 = -2 * 2 * √(6) / 25,

so we finally have that sin(2𝜃) = -4*√(6) / 25,

and

(b) cos(2𝜃) = cos²(𝜃) - sin²(𝜃) = [ 1 - sin²(𝜃) ] - sin²(𝜃) = 1 - 2 sin²(𝜃) = 1 - 2 (-1/5)² = 25/25 - 2*1/25 = 23/25.