Question

find the $sin 165^circ$ using the half-angle formula. rationalize the denominator if necessary.

half-angle formulas
$sinleft(\frac{\theta}{2}
ight) = pm sqrt{\frac{1 - cos(\theta)}{2}}$
$cosleft(\frac{\theta}{2}
ight) = pm sqrt{\frac{1 + cos(\theta)}{2}}$

$sin 165^circ = \frac{sqrt{?} - sqrt{phantom{x}}}{phantom{x}}$

Explanation:

Step1: Determine θ for half - angle

We know that \(165^{\circ}=\frac{330^{\circ}}{2}\), so \(\theta = 330^{\circ}\) and we use the half - angle formula for sine: \(\sin(\frac{\theta}{2})=\pm\sqrt{\frac{1 - \cos\theta}{2}}\). Since \(165^{\circ}\) is in the second quadrant where sine is positive, we take the positive sign. So \(\sin(165^{\circ})=\sin(\frac{330^{\circ}}{2})=\sqrt{\frac{1-\cos(330^{\circ})}{2}}\).

Step2: Find \(\cos(330^{\circ})\)

The angle \(330^{\circ}\) is in the fourth quadrant, and \(\cos(330^{\circ})=\cos(- 30^{\circ}+360^{\circ})=\cos(30^{\circ})=\frac{\sqrt{3}}{2}\) (because cosine is even and \(\cos(360^{\circ}- \alpha)=\cos\alpha\)).

Step3: Substitute \(\cos(330^{\circ})\) into the formula

Substitute \(\cos(330^{\circ})=\frac{\sqrt{3}}{2}\) into \(\sqrt{\frac{1 - \cos(330^{\circ})}{2}}\):
\[

$$\begin{align*} \sin(165^{\circ})&=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{2 - \sqrt{3}}{4}}\\ &=\frac{\sqrt{2-\sqrt{3}}}{2} \end{align*}$$

\]
To rationalize or simplify the numerator, we know that \(\sqrt{2-\sqrt{3}}=\frac{\sqrt{4 - 2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}\) (since \((\sqrt{3}-1)^2 = 3 - 2\sqrt{3}+1=4 - 2\sqrt{3}\)). Then \(\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{2}\) (after rationalizing the denominator of \(\frac{\sqrt{3}-1}{\sqrt{2}}\) by multiplying numerator and denominator by \(\sqrt{2}\)). But from the given form \(\sin165^{\circ}=\frac{\sqrt{[?]-\sqrt{\square}}}{2}\), comparing with \(\frac{\sqrt{2 - \sqrt{3}}}{2}\), we can see that the first square is \(2\) and the second square is \(3\) (or if we consider the simplified form \(\frac{\sqrt{6}-\sqrt{2}}{2}\), but the form in the problem is \(\frac{\sqrt{[?]-\sqrt{\square}}}{2}\), so from \(\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2 - \sqrt{3}}}{2}\), the value inside the outer square root for the numerator is \(2-\sqrt{3}\), so the first? is \(2\) and the \(\square\) is \(3\).

Answer:

\(\sin165^{\circ}=\frac{\sqrt{2 - \sqrt{3}}}{2}\) (or in the form given in the problem, the numerator is \(\sqrt{2-\sqrt{3}}\), so the first blank is \(2\) and the second blank is \(3\), and the denominator is \(2\))