1 Answer

Kernel James · Answer 1 · 2023-09-01T06:24:17+0000

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)

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