You can find the exact value of sin(15 degrees) using both the sum or difference formula and the half-angle identity.
**Using the Sum or Difference Formula:**
The sum or difference formula for sine is:
\[ \sin(A ± B) = \sin(A) * \cos(B) ± \cos(A) * \sin(B) \]
In this case, you can use the difference formula to express sin(15) as a combination of known values. Start with \( A = 45^\circ \) and \( B = 30^\circ \) because \( 45^\circ - 30^\circ = 15^\circ \).
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ) * \cos(30^\circ) - \cos(45^\circ) * \sin(30^\circ) \]
Now, we know that \( \sin(45^\circ) = \cos(45^\circ) = \frac{1}{\sqrt{2}} \) and \( \sin(30^\circ) = \frac{1}{2} \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).
So, substituting these values:
\[ \sin(15^\circ) = \left(\frac{1}{\sqrt{2}}\right) * \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) * \left(\frac{1}{2}\right) \]
Now, simplify this expression.
\[ \sin(15^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]
**Using the Half-Angle Identity:**
The half-angle identity for sine is:
\[ \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \]
In this case, \( \theta = 30^\circ \), and you want to find \( \sin\left(\frac{30^\circ}{2}\right) = \sin(15^\circ) \).
So, apply the half-angle identity:
\[ \sin(15^\circ) = \pm \sqrt{\frac{1 - \cos(30^\circ)}{2}} \]
Now, we know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so:
\[ \sin(15^\circ) = \pm \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \]
Simplify this expression:
\[ \sin(15^\circ) = \pm \sqrt{\frac{2 - \sqrt{3}}{4}} = \pm \frac{\sqrt{2 - \sqrt{3}}}{2} \]
Now, we have the exact value for sin(15 degrees) from both methods:
Using the sum or difference formula: \( \sin(15^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
Using the half-angle identity: \( \sin(15^\circ) = \pm \frac{\sqrt{2 - \sqrt{3}}}{2} \)
To show that these two answers are the same, we can compare them. Note that the plus-minus sign in the half-angle identity can be resolved by considering the quadrant in which the angle falls. In this case, 15 degrees is in the first quadrant, so it's positive. Therefore, both expressions yield the same positive value.
So, \( \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{2 - \sqrt{3}}}{2} \).
To confirm their equality, you can square both sides and simplify:
\[ \left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right)^2 = \left(\frac{\sqrt{2 - \sqrt{3}}}{2}\right)^2 \]
Simplify further, and you'll find that both sides are equal.