By the Pythagorean theorem,
[tex]\sin^2 t +\cos^2 t = 1 \\ \\ \sin^2 t +(-3/4)^2 = 1 \\ \\ \sin^2 t =7/16 \\ \\ \sin t=-\frac{\sqrt{7}}{4}[/tex]
(I took the negative case because of the range of values of t)
Using the cosine double angle formula,
[tex]\cos 2t=2\cos^2 t -1 =2(-3/4)^2 - 1 =1/8[/tex]
Using the sine double angle formula,
[tex]\sin 2t=2\left(-\frac{3}{4} \right)\left(-\frac{\sqrt{7}}{4} \right)=\frac{3\sqrt{7}}{8}[/tex]
Using the cosine half-angle formula,
[tex]\cos \frac{t}{2}= -\frac{\sqrt{1+\cos A}}{2}=-\frac{\sqrt{1+\frac{3\sqrt{7}}{8}}}{2}[/tex]
(I took the negative case because of the range of values of t)
Using the sine half-angle formula,
[tex]\sin \frac{t}{2}=\frac{\sqrt{1-\cos A}}{2}=\frac{\sqrt{1-\frac{3\sqrt{7}}{8}}}{2}[/tex]
(I took the positive case because of the range of values of t)
