# How do you find the exact values of sin 15 degrees using the half angle formula?

Jul 21, 2015

I found: sin(15°)=0.258

#### Explanation:

Using the Half Angle Formula:
$\textcolor{red}{{\sin}^{2} \left(x\right) = \frac{1}{2} \left[1 - \cos \left(2 x\right)\right]}$

with x=15° and 2x=30°

you get:

sin^2(15°)=1/2[1-cos(30°)]

knowing that: cos(30°)=sqrt(3)/2:

sin^2(15°)=1/2[1-sqrt(3)/2]
sin^2(15°)=(2-sqrt(3))/4=0.067

So:
sin(15°)=+-sqrt(0.067)=+-0.258
We choose the positive one.

May 14, 2017

$\sin \left({15}^{\circ}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$
See below.

#### Explanation:

Find exact value of $\sin \left({15}^{\circ}\right)$ with half-angle formula.

Consider the half-angle formula for sine: $\sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos x}{2}}$

Since we know that 15 is half of 30, we can plug ${30}^{\circ}$ in as $\theta$ and simplify:
$\sin \left({15}^{\circ}\right) = \sin \left({30}^{\circ} / 2\right) = \sqrt{\frac{1 - \cos \left({30}^{\circ}\right)}{2}}$
$= \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$
$= \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}$
$= \sqrt{\frac{2 - \sqrt{3}}{4}}$
or see slightly more advanced method to remove nested root (at the bottom)
$= \frac{\sqrt{2 - \sqrt{3}}}{2}$
which is our answer, but has a nested root
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Let us multiply both the numerator and denominator inside the square root by $2$:
$= \sqrt{\frac{2 - \sqrt{3}}{4} \cdot \frac{2}{2}}$
$= \sqrt{\frac{4 - 2 \sqrt{3}}{8}}$

Now we can write $4 - 2 \sqrt{3}$ as a square in the numerator:
$= \sqrt{{\left(\sqrt{3} - 1\right)}^{2} / 8}$

We can take out a $\sqrt{3} - 1$ from the numerator and a $2$ from the denominator:
$= \left(\frac{\sqrt{3} - 1}{2}\right) \cdot \frac{1}{\sqrt{2}}$
$= \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

We can rationalize the denominator:
$= \frac{\sqrt{2} \left(\sqrt{3} - 1\right)}{4}$
$= \frac{\sqrt{6} - \sqrt{2}}{4}$

which is our answer without a nested root.

Aug 5, 2018

$\frac{\sqrt{6} - \sqrt{2}}{4}$

#### Explanation:

Use

$\sin \left(A - B\right) = \sin A \cdot \cos B - S \in B \cdot \cos A$

So

$\sin 15 = \sin \left(45 - 30\right) = \sin 45 \cdot \cos 30 - \sin 30 \cdot \cos 45$

We know

$\sin 45 = \cos 45 = \frac{\sqrt{2}}{2}$

We also know

$\cos 30 = \frac{\sqrt{3}}{2}$

$\sin 30 = \frac{1}{2}$

Plugging in the values, we get

$\sin 15 = \frac{\sqrt{2}}{2} \cdot \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)$

Simplifying, we get

$\frac{\sqrt{6} - \sqrt{2}}{4}$