Calculate Sin(75°) Using Half-Angle Formula

Let's dive into calculating the value of $\sin 75^{\circ}$ using the half-angle formula. This is a classic problem in trigonometry that helps illustrate the power and utility of these formulas. We'll break it down step by step, ensuring you understand not just the mechanics but also the underlying concepts. So, grab your calculators (or your trusty pen and paper), and let's get started!

The half-angle formulas are derived from the double-angle formulas and are incredibly useful when you need to find trigonometric functions of angles that are half of a more common or easily calculated angle. In our case, $75^{\circ}$ is half of $150^{\circ}$, and we know the trigonometric values for $150^{\circ}$ (or at least, we can easily derive them). The half-angle formula for sine is given by:

$$\sin \left(\frac{\theta}{2}\right)= \pm \sqrt{\frac{1-\cos (\theta)}{2}}$$

The plus or minus sign in front of the square root might seem a bit confusing at first, but it simply means we need to consider the quadrant in which the angle $\frac{\theta}{2}$ lies to determine the correct sign. Since $75^{\circ}$ is in the first quadrant, where sine is positive, we'll take the positive square root.

Applying the Half-Angle Formula

Here’s how we can apply the half-angle formula to find $\sin 75^{\circ}$:

1. Identify $\theta$: We want to find $\sin 75^{\circ}$, so we set $\frac{\theta}{2} = 75^{\circ}$. This means $\theta = 150^{\circ}$.

2. Find $\cos \theta$: We need to find the cosine of $150^{\circ}$. Recall that $150^{\circ}$ is in the second quadrant, where cosine is negative. We can express $150^{\circ}$ as $180^{\circ} - 30^{\circ}$. Therefore, $\cos 150^{\circ} = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$.

3. Plug into the Formula: Now we substitute $\cos 150^{\circ}$ into the half-angle formula:

   $$\sin 75^{\circ} = \sqrt{\frac{1 - \cos 150^{\circ}}{2}} = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{2}\right)}{2}}$$

4. Simplify: Let's simplify the expression step by step:

   $$\sin 75^{\circ} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}}$$

   $$\sin 75^{\circ} = \frac{\sqrt{2 + \sqrt{3}}}{2}$$

Rationalizing the Denominator (If Necessary)

In this case, the denominator is already rational (it's just 2), so we don't need to rationalize it. However, the expression under the square root can be further simplified. While not strictly necessary, it's often good practice to simplify as much as possible.

To simplify $\sqrt{2 + \sqrt{3}}$, we can try to express $2 + \sqrt{3}$ as a perfect square. Let's assume that:

$$2 + \sqrt{3} = (a + b)^2 = a^2 + 2ab + b^2$$

We want to find $a$ and $b$ such that $a^2 + b^2 = 2$ and $2ab = \sqrt{3}$. A bit of algebraic manipulation (and perhaps some educated guessing) leads us to:

$$a = \frac{\sqrt{6}}{2}, \quad b = \frac{\sqrt{2}}{2}$$

Let's check if this works:

$$a^2 + b^2 = \left(\frac{\sqrt{6}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{6}{4} + \frac{2}{4} = \frac{8}{4} = 2$$

$$2ab = 2 \cdot \frac{\sqrt{6}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{2\sqrt{12}}{4} = \frac{\sqrt{4 \cdot 3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$$

It works! So, we can write:

$$\sqrt{2 + \sqrt{3}} = \frac{\sqrt{6} + \sqrt{2}}{2}$$

Therefore,

$$\sin 75^{\circ} = \frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\frac{\sqrt{6} + \sqrt{2}}{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Final Answer

So, the final answer is:

$$\sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Alternative Approach: Sum-to-Product Formula

While we used the half-angle formula as requested, it's worth noting that you can also find $\sin 75^{\circ}$ using the sum of angles formula:

$$\sin (A + B) = \sin A \cos B + \cos A \sin B$$

We can express $75^{\circ}$ as $45^{\circ} + 30^{\circ}$. We know the sine and cosine values for both $45^{\circ}$ and $30^{\circ}$:

• $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$
• $\cos 45^{\circ} = \frac{\sqrt{2}}{2}$
• $\sin 30^{\circ} = \frac{1}{2}$
• $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$

Plugging these values into the sum of angles formula:

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

$$\sin 75^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

As you can see, we arrive at the same answer using a different method!

Key Takeaways

• The half-angle formulas are powerful tools for finding trigonometric functions of angles that are half of known angles.
• Always consider the quadrant of the angle to determine the correct sign when using half-angle formulas.
• Simplifying radical expressions can often lead to a more elegant final answer.
• There are often multiple ways to solve a trigonometry problem, so explore different approaches to deepen your understanding.

Practice Problems

To solidify your understanding, try these practice problems:

1. Find $\cos 15^{\circ}$ using the half-angle formula.
2. Find $\tan 22.5^{\circ}$ using the half-angle formula.
3. Verify your answers using other trigonometric identities or formulas.

Conclusion

Calculating $\sin 75^{\circ}$ using the half-angle formula is a great exercise in applying trigonometric identities and simplifying radical expressions. Remember, the key is to break down the problem into smaller, manageable steps and to pay attention to the details. With practice, you'll become more comfortable and confident in your ability to tackle these types of problems. So, keep practicing, and happy calculating!

By understanding and applying the half-angle formula, we've successfully found the value of $\sin 75^{\circ}$. This showcases the versatility of trigonometric identities and their importance in solving a variety of mathematical problems. Keep exploring and practicing, and you'll master these concepts in no time!