Question

find the sin 15° using the half-angle formula. rationalize the denominator if necessary. half-angle formulas sin(θ/2) = ±√(1 - cos(θ))/2 cos(θ/2) = ±√(1 + cos(θ))/2 sin 15° = √? - √ /

Explanation:

Step1: Identify θ for half - angle

We know that \(15^{\circ}=\frac{30^{\circ}}{2}\), so \(\theta = 30^{\circ}\) and we will use the half - angle formula for sine: \(\sin(\frac{\theta}{2})=\pm\sqrt{\frac{1 - \cos\theta}{2}}\). Since \(15^{\circ}\) is in the first quadrant, the sine value is positive, so we use the positive form \(\sin(15^{\circ})=\sqrt{\frac{1-\cos(30^{\circ})}{2}}\).

Step2: Substitute \(\cos(30^{\circ})\) value

We know that \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\). Substitute this into the formula:
\[ \begin{align*} \sin(15^{\circ})&=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{2-\sqrt{3}}{4}}\\ &=\frac{\sqrt{2 - \sqrt{3}}}{\sqrt{4}}\\ &=\frac{\sqrt{2-\sqrt{3}}}{2} \end{align*} \]
We can also rationalize or rewrite \(\sqrt{2 - \sqrt{3}}\) in a more expanded form. 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 \(\sqrt{3}-1>0\)). But if we want to write it in the form \(\frac{\sqrt{?}-\sqrt{?}}{?}\), we can square \(\frac{\sqrt{a}-\sqrt{b}}{c}\) and equate it to \(\frac{2 - \sqrt{3}}{4}\).

Let \(\frac{\sqrt{a}-\sqrt{b}}{c}=\sqrt{\frac{2 - \sqrt{3}}{4}}=\frac{\sqrt{2 - \sqrt{3}}}{2}\). Squaring both sides: \(\frac{a + b-2\sqrt{ab}}{c^{2}}=\frac{2-\sqrt{3}}{4}\).

We want \(a + b = 2\) and \(2\sqrt{ab}=\sqrt{3}\), or \(4ab = 3\). Solving the system \(\begin{cases}a + b=2\\ab=\frac{3}{4}\end{cases}\), the quadratic equation \(x^{2}-2x+\frac{3}{4}=0\), with solutions \(x=\frac{2\pm\sqrt{4 - 3}}{2}=\frac{2\pm1}{2}\), so \(x = \frac{3}{2}\) or \(x=\frac{1}{2}\).

So \(\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-\sqrt{1}}{\sqrt{2}}\) (after some algebraic manipulation). Then \(\sin(15^{\circ})=\frac{\sqrt{3}-\sqrt{1}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{3}-\sqrt{1}}{2}\) (wait, no, let's do it correctly).

Wait, going back, when we have \(\sin(15^{\circ})=\sqrt{\frac{2 - \sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}\). We can also note that \(2-\sqrt{3}=\frac{4 - 2\sqrt{3}}{2}=\frac{(\sqrt{3}-1)^2}{2}\), so \(\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}\), then \(\sin(15^{\circ})=\frac{\sqrt{3}-1}{2\sqrt{2}}\times\sqrt{2}\times\sqrt{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\)? Wait, no, I made a mistake earlier.

Wait, let's start over. The half - angle formula:

\(\sin(15^{\circ})=\sin(\frac{30^{\circ}}{2})=\sqrt{\frac{1 - \cos30^{\circ}}{2}}\)

\(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), so:

\[ \begin{align*} \sin(15^{\circ})&=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{2-\sqrt{3}}{4}}\\ &=\frac{\sqrt{2 - \sqrt{3}}}{2} \end{align*} \]

Now, we can rewrite \(\sqrt{2-\sqrt{3}}\) as \(\frac{\sqrt{4 - 2\sqrt{3}}}{\sqrt{2}}\). And \(4-2\sqrt{3}=(\sqrt{3}-1)^2\), so \(\sqrt{4 - 2\sqrt{3}}=\sqrt{3}-1\). Then \(\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}\), so \(\sin(15^{\circ})=\frac{\sqrt{3}-1}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\). But if we want to write it in the form \(\frac{\sqrt{?}-\sqrt{?}}{?}\), we can see that \(\sqrt{6}-\sqrt{2}\) in the numerator and \(4\) in the denominator? Wait, no, let's check the form given in the problem \(\sin15^{\circ}=\frac{\sqrt{?}-\sqrt{?}}{?}\).

We know that \(\sin(15^{\circ})=\frac{\sqrt{6}-\sqrt{2}}{4}\). Let's verify: \((\sqrt{6}-\sqrt{2})^2=6 + 2-2\sqrt{12}=8 - 4\sqrt{3}\), and \((2\sqrt{2})^2 = 8\), \(\frac{8 - 4\sqrt{3}}{8}=\frac{2-\sqrt{3}}{2}\), no, wait. Wait, the correct form from the half - angle formula steps:

We have \(\sin(15^{\circ})=\sqrt{\frac{1-\cos30^{\circ}}{2}}=\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}\). If we rationalize or express \(\sqrt{2-\sqrt{3}}\) as \(\frac{\sqrt{4 - 2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}\) (since \(\sqrt{3}-1>0\)). Then \(\sin(15^{\circ})=\frac{\sqrt{3}-1}{2\sqrt{2}}\times\sqrt{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\).

Wait, the problem's form is \(\sin15^{\circ}=\frac{\sqrt{?}-\sqrt{?}}{?}\). So comparing \(\frac{\sqrt{6}-\sqrt{2}}{4}\), we have the numerator as \(\sqrt{6}-\sqrt{2}\) and the denominator as \(4\). But also, from the half - angle formula steps, when we had \(\sin(15^{\circ})=\sqrt{\frac{2 - \sqrt{3}}{4}}\), if we consider the expression inside the square root \(\frac{2-\sqrt{3}}{4}=\frac{4 - 2\sqrt{3}}{8}=\frac{(\sqrt{3}-1)^2}{8}\)? No, let's go back to the basic half - angle formula application.

The half - angle formula for sine: \(\sin(\frac{\theta}{2})=\sqrt{\frac{1 - \cos\theta}{2}}\) (for \(\frac{\theta}{2}\) in first quadrant). Here \(\theta = 30^{\circ}\), so:

\(\sin(15^{\circ})=\sqrt{\frac{1-\cos(30^{\circ})}{2}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2 - \sqrt{3}}{4}}=\frac{\sqrt{2 - \sqrt{3}}}{2}\)

We can also write \(\sqrt{2-\sqrt{3}}\) as \(\frac{\sqrt{4 - 2\sqrt{3}}}{\sqrt{2}}\), and \(4 - 2\sqrt{3}=(\sqrt{3}-1)^2\), so \(\sqrt{4 - 2\sqrt{3}}=\sqrt{3}-1\), then \(\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}\), so \(\sin(15^{\circ})=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\) (after rationalizing the denominator of \(\frac{\sqrt{3}-1}{2\sqrt{2}}\) by multiplying numerator and denominator by \(\sqrt{2}\)).

So in the form \(\frac{\sqrt{?}-\sqrt{?}}{?}\), the numerator is \(\sqrt{6}-\sqrt{2}\) and the denominator is \(4\), or if we consider the initial half - angle steps, when we have \(\sin(15^{\circ})=\sqrt{\frac{2-\sqrt{3}}{4}}\), and if we want to write it as \(\frac{\sqrt{a}-\sqrt{b}}{c}\), by squaring \(\frac{\sqrt{a}-\sqrt{b}}{c}\) we get \(\frac{a + b-2\sqrt{ab}}{c^{2}}=\frac{2-\sqrt{3}}{4}\). Let \(c = 2\), then \(a + b-2\sqrt{ab}=2-\sqrt{3}\). Let \(a = 2\), \(b=\frac{3}{2}\), no, better to note that \(\frac{\sqrt{6}-\sqrt{2}}{4}\) is the correct simplified form. So the values are: the first square root is \(6\), the second square root is \(2\), and the denominator is \(4\). But also, from the half - angle formula, when \(\theta = 30^{\circ}\), the formula gives us \(\sin(15^{\circ})=\sqrt{\frac{1-\cos30^{\circ}}{2}}\), and if we expand \(\sqrt{\frac{2 - \sqrt{3}}{4}}\) as \(\frac{\sqrt{2-\sqrt{3}}}{2}\), and \(\sqrt{2-\sqrt{3}}\) can be written as \(\frac{\sqrt{4 - 2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}\), so \(\sin(15^{\circ})=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\) (after rationalizing).

Answer:

\(\sin15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}\), so the values in the boxes are \(6\), \(2\), and \(4\) (from left to right in the numerator square roots and the denominator). So the answer is \(\frac{\sqrt{6}-\sqrt{2}}{4}\)