# How do you find the exact value of sin 105 degrees?

Oct 25, 2015

Find exact value of sin (105)

Ans: $\left(\frac{\sqrt{2 + \sqrt{3}}}{2}\right)$

#### Explanation:

sin (105) = sin (15 + 90) = cos 15.
First find (cos 15). Call cos 15 = cos x
Apply the trig identity:$\cos 2 x = 2 {\cos}^{2} x - 1.$
cos 2x = cos (30) $= \frac{\sqrt{3}}{2} = 2 {\cos}^{2} x - 1$
2cos^2 x = 1 + sqrt3/2 = (2 + sqrt3)/2
cos^2 x = (2 + sqrt3)/4
cos x = cos 15 = (sqrt(2 + sqrt3)/2. (since cos 15 is positive)

$\sin \left(105\right) = \cos \left(15\right) = \frac{\sqrt{2 + \sqrt{3}}}{2.}$
Check by calculator.
sin (105) = cos 15 = 0.97
$\frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{1.93}{2} = 0.97 .$ OK

Feb 10, 2016

Use $\sin 105 = \sin \left(60 + 45\right) = \sin 60 \cos 45 + \cos 60 \sin 45$
=(sqrt3/2)(1/sqrt2)+(1/2)(1/sqrt2)=sqrt2/4((sqrt3+1)=0.9656 nearly.

#### Explanation:

sin 45, cos 45 and sin 60 are irrational. So, the answer is a surd.

Jan 29, 2017

$\sin {105}^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$

#### Explanation:

We know $\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$

Hence $\sin {105}^{\circ}$

= $\sin \left({60}^{\circ} + {45}^{\circ}\right)$

= $\sin {60}^{\circ} \cos {45}^{\circ} + \cos {60}^{\circ} \sin {45}^{\circ}$

= $\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}}$

= $\frac{\sqrt{3} + 1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

= $\frac{\sqrt{6} + \sqrt{2}}{4}$