How would I find this, given the equation?

Question

Given: sin𝛼 =3/8,𝜋/2 < 𝛼 < 𝜋;cos𝛽 =12/13,3𝜋2 < 𝛽 < 2𝜋

How would I find?

1) sin 2𝛽 =

and

2) sin𝛽/2 =

I drew a triange to find sin,cos,tan,etc however I don’t know what to do from there? Thank you

Answer 1

Hi Christoforos,

1) You need to use the double-angle identity. For sine, this is sin(2𝛽) = 2sin(𝛽)cos(𝛽). The problem gives us that cos(𝛽) = 12/13. To find sin(𝛽), you have to use sin²(𝛽) + cos²(𝛽) = 1, so sin(𝛽) = √(1 - cos²(𝛽)) = √(1 - 144/169) = 5/13 (we'll figure out whether + or - later). So, this means sin(2𝛽) = 2(5/13)(12/13) = 120/169. Since 𝛽 lies between 3π/2 and 2π, 2𝛽 lies between 3pi and 4pi, which is the same as quadrants 3 and 4. In these quadrants, sine is negative, so our final answer is -120/169.

2) This needs the half-angle identity, which for sine is sin(𝛽/2) = √((1 - cos(𝛽)) / 2). cos(𝛽) = 12/13, so sin(𝛽/2) = √((1 - 12/13) / 2) = √(1/26) = √(26)/26. The range of 𝛽/2 is between 3π/4 and π, which is the second quadrant. Sine is still positive here, so the answer remains as √(26)/26.

Hope this helps!

Akshat Y.