Question

Please help!
The picture below shows the question I am stuck on!
Thank you in advance!

Answer 1

By the Pythagorean theorem,

`\sin^2 t +\cos^2 t = 1 \\ \\ \sin^2 t +(-3/4)^2 = 1 \\ \\ \sin^2 t =7/16 \\ \\ \sin t=-(√(7))/(4)`

(I took the negative case because of the range of values of t)

Using the cosine double angle formula,

`\cos 2t=2\cos^2 t -1 =2(-3/4)^2 - 1 =1/8`

Using the sine double angle formula,

`\sin 2t=2\left(-(3)/(4) \right)\left(-(√(7))/(4) \right)=(3√(7))/(8)`

Using the cosine half-angle formula,

`\cos (t)/(2)= -(√(1+\cos A))/(2)=-\frac{\sqrt{1+(3√(7))/(8)}}{2}`

(I took the negative case because of the range of values of t)

Using the sine half-angle formula,

`\sin (t)/(2)=(√(1-\cos A))/(2)=\frac{\sqrt{1-(3√(7))/(8)}}{2}`

(I took the positive case because of the range of values of t)