Certainly! Let's verify the trigonometric identity [tex]\(\sin (A + B) = \sin A \cos B + \cos A \sin B\)[/tex] using the given angles [tex]\(A = 60^\circ\)[/tex] and [tex]\(B = 30^\circ\)[/tex].
### Step-by-Step Solution:
1. Calculate [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
- Given [tex]\(A = 60^\circ\)[/tex]:
[tex]\[
\sin A = \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386
\][/tex]
[tex]\[
\cos A = \cos 60^\circ = \frac{1}{2} \approx 0.5000000000000001
\][/tex]
2. Calculate [tex]\(\sin B\)[/tex] and [tex]\(\cos B\)[/tex]:
- Given [tex]\(B = 30^\circ\)[/tex]:
[tex]\[
\sin B = \sin 30^\circ = \frac{1}{2} \approx 0.49999999999999994
\][/tex]
[tex]\[
\cos B = \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844387
\][/tex]
3. Calculate [tex]\(\sin (A + B)\)[/tex]:
- Here, [tex]\(A + B = 60^\circ + 30^\circ = 90^\circ\)[/tex]:
[tex]\[
\sin (A + B) = \sin 90^\circ = 1
\][/tex]
4. Compute the right-hand side of the identity [tex]\(\sin A \cos B + \cos A \sin B\)[/tex]:
[tex]\[
\sin A \cos B + \cos A \sin B = \left( \frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{1}{2} \right)
\][/tex]
This simplifies to:
[tex]\[
= \frac{3}{4} + \frac{1}{4} = 1
\][/tex]
5. Compare the left-hand side and the right-hand side:
- The left-hand side [tex]\(\sin (A + B)\)[/tex] is 1.
- The right-hand side [tex]\(\sin A \cos B + \cos A \sin B\)[/tex] is 1.
Since both sides are equal, we have successfully verified the identity:
[tex]\[
\sin (A + B) = \sin A \cos B + \cos A \sin B
\][/tex]
Thus, [tex]\(\sin (60^\circ + 30^\circ) = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\)[/tex] holds true.
