12. Recall the half-angle identity,[tex]\sin^2\dfrac y2=\dfrac{1-\cos y}2\implies\sin\dfrac y2=\sqrt{\dfrac{1-\cos y}2}[/tex]where we take the positive square root because [tex]y[/tex] is an angle in a right triangle, which means [tex]0^\circ<y<90^\circ[/tex], so [tex]0^\circ<\dfrac y2<45^\circ[/tex]. For such an angle, it's always the case that [tex]\sin\dfrac y2>0[/tex].Use the Pythagorean theorem to find the length of the hypotenuse:[tex]\sqrt{5^2+12^2}=\sqrt{169}=13[/tex]Then[tex]\sin\dfrac y2=\sqrt{\dfrac{1-\frac5{13}}2}=\sqrt{\dfrac4{13}}=\dfrac2{\sqrt{13}}[/tex]###13. [tex]x[/tex] is in quadrant II, which means [tex]90^\circ<x<180^\circ[/tex], so [tex]45^\circ<\dfrac x2<90^\circ[/tex]. In other words, [tex]\dfrac x2[/tex] is in quadrant I, so [tex]\sin\dfrac x2>0[/tex]. From the half-angle identity we get[tex]\sin\dfrac x2=\sqrt{\dfrac{1-\cos x}2}=\sqrt{\dfrac23}[/tex]###14. Simplification follows from the definitions of each function:[tex]\sec x=\dfrac1{\cos x}[/tex][tex]\tan x=\dfrac{\sin x}{\cos x}[/tex][tex]\csc x=\dfrac1{\sin x}[/tex]So we have[tex]\sec x\tan x\cos x\csc x=\dfrac{\sin x\cos x}{\cos^2x\sin x}=\dfrac1{\cos x}[/tex]###15. Use the Pythagorean identity:[tex]\cos^2x+\sin^2x=1\implies\dfrac{\cos^2x}{\cos^2x}+\dfrac{\sin^2x}{\cos^2x}=\dfrac1{\cos^2x}\implies1+\tan^2x=\sec^2x[/tex]Then[tex]\tan^3x+\tan x=\tan x(\tan^2x+1)=\tan x\sec^2x[/tex]