Solving for the Value of (sin $75^\circ$ + sin $15^\circ$)

The question asks us to find the exact value of the sum of the sine of $75$ degrees and the sine of $15$ degrees. We can solve this using trigonometric identities.

Method 1: Using the Sum-to-Product Formula

One efficient way to solve this is by using the trigonometric sum-to-product identity:

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

In this problem, let $A = 75^\circ$ and $B = 15^\circ$. Applying the formula:

• Calculate $\frac{A+B}{2}$: $$ \frac{75^\circ + 15^\circ}{2} = \frac{90^\circ}{2} = 45^\circ $$
• Calculate $\frac{A-B}{2}$: $$ \frac{75^\circ - 15^\circ}{2} = \frac{60^\circ}{2} = 30^\circ $$
• Substitute these values back into the formula: $$ \sin 75^\circ + \sin 15^\circ = 2 \sin(45^\circ) \cos(30^\circ) $$

Now, we use the known values of $\sin 45^\circ$ and $\cos 30^\circ$:

• $ \sin 45^\circ = \frac{\sqrt{2}}{2} $
• $ \cos 30^\circ = \frac{\sqrt{3}}{2} $

Substitute these values:

$$ 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} = \frac{2 \times \sqrt{2} \times \sqrt{3}}{2 \times 2} = \frac{2\sqrt{6}}{4} $$

Simplify the expression:

$$ \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2} $$

Method 2: Using Angle Addition and Subtraction Formulas

Alternatively, we can calculate $\sin 75^\circ$ and $\sin 15^\circ$ individually and then add them.

We can express $75^\circ$ as $45^\circ + 30^\circ$ and $15^\circ$ as $45^\circ - 30^\circ$.

• Calculating $\sin 75^\circ$:

  Using the sine addition formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

  $$ \sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ $$

  $$ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} $$

• Calculating $\sin 15^\circ$:

  Using the sine subtraction formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:

  $$ \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ $$

  $$ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} $$

• Adding the results:

  $$ \sin 75^\circ + \sin 15^\circ = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) + \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) $$

  $$ = \frac{(\sqrt{6} + \sqrt{2}) + (\sqrt{6} - \sqrt{2})}{4} = \frac{\sqrt{6} + \sqrt{2} + \sqrt{6} - \sqrt{2}}{4} = \frac{2\sqrt{6}}{4} $$

  $$ = \frac{\sqrt{6}}{2} $$

Conclusion

Both methods confirm that the value of $\sin 75^\circ + \sin 15^\circ$ is $\frac{\sqrt{6}}{2}$. This matches option 1.