Question 825715
If sinθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ.
F. -24/25
G. 8/5 (I chose this one and it was wrong)
H.-12/25
J. 24/25
Sorry I found the typo I submitted in one of my previous questions. So sorry about that.

<pre>
sin &#952; = {{{4/5}}} = {{{O/H}}}
Since the opposite side (O) is 4 and the hypotenuse (H)is 5, the adjacent side (A) MUST = 3,
as this represents a 3-4-5 special triangle

Since &#952; terminates in the 2nd quadrant, where cos is < 0 (negative), we get:
cos &#952; = {{{- A/H = - 3/5}}} 

sin 2 &#952; = 2 sin &#952; cos &#952; (double-angle identity)
Therefore, sin 2 &#952; = {{{2(4/5)(- 3/5)}}}____{{{-(2(4)(3))/(5)(5)}}}____{{{highlight_green(- 24/25)}}} (CHOICE H)