Given tan alpha 12 sin beta 23 and both alpha beta are in

Given tan alpha 1/2, sin beta = 2/3, and both alpha & beta are in QI find: sin(alpha + beta) cos(2 beta)

Solution

tan = 1/2

sec = 5/2

cos = 2/5

sin = 1/5

sin = 2/3

cos = 5/3

Now sin( + ) =  sin()cos() + cos()sin() = (1/5)*(5/3) + (2/5)*(2/3) = (1/3) + (4/35) = (1/3)*[1 + (4/5)]

And cos(2) = 2cos²() – 1 = 2*5/9 - 1 = (10/9) - 1 = 1/9