Given that theda is in Quadrant II and sin(theda) = 12/13, find sin(theda/2), cos(theda/2)...?

Question

Complete question:
Given that theda is in Quadrant II and sin(theda) = 12/13, find sin(theda/2), cos(theda/2), sin(2 theda), and cos(2 theda)

Once again stuck on a pre-calc question for my review, showing your work would help to prepare me for the test, and any insight would be helpful. And as always, thanks in advance for anyone reading this and trying to help out.

Answer 1

These ratios are based on the 5-12-13 triangle, but in quadrant 2, cos and tan are negative.
sin(theda) = 12/13 (given)
cos(theda) = -5/13
tan(theda) = -12/5

If t = tan(theda/2), then:
2t / (1 + t^2) = sin(theda) = 12/13
26t = 12(1 + t^2)
= 12 + 12t^2
6t^2 - 13t + 6 = 0
(3t - 2)(2t - 3) = 0
t = 2/3 or t = 3/2
Only t = 3/2 puts theda in quadrant 2, and theda/2 is in quadrant 1.
The hypotenuse of a triangle with sides 3 and 2 is sqrt(3^2 + 2^2) = sqrt(13).
Hence sin(theda / 2) = 3/sqrt(13) = 3sqrt(13)/13
and cos(theda / 2) = 2/sqrt(13) = 2sqrt(13)/13

sin(2theda) = 2sin(theda)cos(theda)
= 2 (12/13) (-5/13) = -120 / 169
cos(2theda) = 2cos^2(theda) - 1
= 2 (25/169) - 1
= 50 / 169.

Answer 2

Your question was a little confusing. Okay, Sin=y/r Cos=x/r

your y is 12 and your r is 13. if you use the Pythagorean theorem you find that x is 5. so, just do the sin/2 and the cos/2. then do the cos *2. and then the sin*2.

i think. hopefully that's what you were asking for

Answer 3

Let theta = x
sin x =12/13 thus cos x =-5/13 [5=sqrt13^2-12^2) and cos is negative in quadrant II].
sin 2x = 2sinxcosx =2*(12/13)(-5/13) = -.71006
cos 2x = 1-2sin^2x =1 -2(12/13)^2 = .70414
sin x/2 = sqrt((1-cosx)/2)) = sqrt((1+5/13)/2) = .83205
cos x/2 = sqrt((1+cosx)/2) = sqrt((1-5/13)/2) =.5547