Finding the Value of sin 75° + sin 15°

This problem asks for the value of the sum of sines of two different angles. We can solve this using a trigonometric sum-to-product identity.

Using the Sum-to-Product Identity for Sine

The relevant trigonometric identity for the sum of two sines is:

\(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)

In this problem, we have \(A = 75°\) and \(B = 15°\).

Step-by-Step Calculation

Let's apply the identity step-by-step:

1. Identify the angles: \(A = 75°\) and \(B = 15°\).
2. Calculate the sum of the angles: \(A + B = 75° + 15° = 90°\).
3. Calculate the average of the angles: \(\frac{A+B}{2} = \frac{90°}{2} = 45°\).
4. Calculate the difference of the angles: \(A - B = 75° - 15° = 60°\).
5. Calculate half of the difference: \(\frac{A-B}{2} = \frac{60°}{2} = 30°\).
6. Substitute these values into the identity:

   \(\sin 75° + \sin 15° = 2 \sin\left(45°\right) \cos\left(30°\right)\)

7. Recall the standard trigonometric values for 45° and 30°:
   • \(\sin 45° = \frac{1}{\sqrt{2}}\)
   • \(\cos 30° = \frac{\sqrt{3}}{2}\)

   Here is a quick reference for some standard angle values:

   Angle (°)	sin	cos	tan
   0	0	1	0
   30	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{\sqrt{3}}\)
   45	\(\frac{1}{\sqrt{2}}\)	\(\frac{1}{\sqrt{2}}\)	1
   60	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)	\(\sqrt{3}\)
   90	1	0	Undefined

8. Substitute the standard values into the expression:

   \(\sin 75° + \sin 15° = 2 \times \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\)

9. Simplify the expression:

   \(\sin 75° + \sin 15° = \frac{2 \times 1 \times \sqrt{3}}{2 \times \sqrt{2}}\)

   \(\sin 75° + \sin 15° = \frac{\sqrt{3}}{\sqrt{2}}\)

10. Match the result with the given options. The expression \(\frac{\sqrt{3}}{\sqrt{2}}\) can also be written as \(\sqrt{\frac{3}{2}}\) by combining the terms under one square root.

Thus, the value of sin 75° + sin 15° is \(\sqrt{\frac{3}{2}}\).

Revision Table: Key Trigonometric Identities

Additional Information: Other Ways to Calculate sin 75° and sin 15°

Although using the sum-to-product identity is efficient here, you could also calculate sin 75° and sin 15° individually using sum/difference identities:

• \(75° = 45° + 30°\)
• \(15° = 45° - 30°\) or \(15° = 60° - 45°\)

Using \(\sin(A+B) = \sin A \cos B + \cos A \sin B\) and \(\sin(A-B) = \sin A \cos B - \cos A \sin B\):

• \(\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30°\)

  \(= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}\)

• \(\sin 15° = \sin(45° - 30°) = \sin 45° \cos 30° - \cos 45° \sin 30°\)

  \(= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}\)

Adding these two values:

\(\sin 75° + \sin 15° = \frac{\sqrt{3}+1}{2\sqrt{2}} + \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{(\sqrt{3}+1) + (\sqrt{3}-1)}{2\sqrt{2}} = \frac{2\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}\)

This confirms the result obtained using the sum-to-product identity. Both methods lead to the same answer, \(\sqrt{\frac{3}{2}}\).