Question

What is the exact value of sin (165) degrees?

Answer 1

Answer:  The required value of sin(165°) is $\dfrac{\sqrt6-\sqrt2}{4}.$

Step-by-step explanation:  We are given to find the exact value of the following trigonometric ratio :

$T_r=\sin (165^\circ).$

We will be using the following trigonometric properties :

$(i)~\sin (180^\circ-x)=\sin x,\\\\(ii)~\sin (A-B)=\sin A \cos B-\cos A sin B.$

We have

$T_r\\\\=\sin (165^\circ)\\\\=\sin (180^\circ-15^\circ)\\\\=\sin 15^\circ\\\\=\sin (45^\circ-30^\circ)\\\\=\sin 45^\circ \cos 30^\circ-\cos 45^\circ \sin 30^\circ\\\\=\dfrac{1}{\sqrt2}.\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2}.\dfrac{1}{2}\\\\\\=\dfrac{\sqrt3-1}{2\sqrt2}\\\\=\dfrac{\sqrt2(\sqrt3-1)}{2\times2}\\\\\\=\dfrac{\sqrt6-\sqrt2}{4}.$

Thus, the required value of sin(165°) is $\dfrac{\sqrt6-\sqrt2}{4}.$

Answer 2

sin 165 = sin(120 + 45) =sin 120 cos 45 + cos 120 sin 45 =
[sqrt(3)/2 * sqrt(2)/2] + (-1/2)(sqrt(2)/2) =
[sqrt(6) - sqrt(2)] / 4