In questions 2A, 2B, and 2C below, find the exact values using the given information:
sin α = 1/6 with α between 0 and π/2
sin β = 3/5 with β between π/2 and π
2A. Find cos (α + β)
2B. Find tan (2β)
2C. Find sin (β/2)
draw the triangles in standard position. You will see that since
sin = y/r
cos = x/r
tan = y/x,
sinα = 1/6
cosα = √35/6
tanα = 1/√35
sinβ = 3/5
cosβ = -4/5
tanβ = 3/-4 = -3/4
Now just crank it out using your sum and double-angle formulas:
cos(α + β) = cosα cosβ - sinα sinβ
= √35/6 * -4/5 - 1/6 * 3/5
= -4√35/30 - 3/30
tan(2β) = 2tanβ/(1-tan^2β)
= 2(-3/4) / (1-9/16)
= (-3/2)/(7/16)
= -24/7
sin(β/2) = √((1-cosβ)/2)
= √((1+4/5)/2)
= √(9/10)
= 3/√10