Finding Sin 75 Degrees Using The Half-Angle Formula

Hey guys! Today, we're diving into the world of trigonometry to figure out the value of $\sin 75^{\circ}$ using the half-angle formula. If you're scratching your head thinking, "What's a half-angle formula?", don't worry, we'll break it down step by step. Trust me, it's way less intimidating than it sounds! So, grab your calculators (or your mental math muscles) and let's get started!

Understanding Half-Angle Formulas

Before we jump into finding $\sin 75^{\circ}$, let's quickly recap what half-angle formulas are all about. These nifty formulas allow us to find the trigonometric functions of an angle that is half of another angle we already know. Think of it as a mathematical magic trick where we transform one angle into another!

The main keywords for half-angle formulas are:

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

These formulas might look a bit scary with their square roots and plus-minus signs, but they're actually quite straightforward once you get the hang of them. The plus-minus sign ($\pm$) is there because the sign of the trigonometric function depends on the quadrant in which the angle $\frac{\theta}{2}$ lies. We'll need to consider this when we apply the formula.

The beauty of these half-angle formulas lies in their ability to simplify complex trigonometric problems. Imagine trying to find the sine or cosine of an angle like 15° or 22.5° without these formulas – it would be a nightmare! But with the half-angle formulas, we can relate these angles to angles we already know, such as 30°, 45°, or 60°, and solve for the trigonometric functions relatively easily. This is why understanding and mastering these formulas is so crucial for anyone venturing deeper into trigonometry and calculus.

To effectively utilize the half-angle formulas, it's essential to have a solid grasp of the unit circle and the trigonometric values of common angles like 0°, 30°, 45°, 60°, and 90°. These angles serve as the foundation for applying the half-angle formulas, as they allow us to relate unknown trigonometric values to known ones. For instance, if we want to find sin(15°), we can recognize that 15° is half of 30°, and we already know the sine and cosine values for 30°. Similarly, for angles like 22.5° (half of 45°) or 67.5° (half of 135°), we can leverage our knowledge of the unit circle and the corresponding trigonometric values to simplify the calculations. Moreover, understanding the quadrant in which the angle lies is crucial for determining the correct sign (positive or negative) when applying the half-angle formulas, as the trigonometric functions have different signs in different quadrants. By mastering these fundamental concepts, you'll be well-equipped to tackle a wide range of trigonometric problems using the half-angle formulas.

Setting Up the Problem for sin 75°

Okay, so we need to find $\sin 75^{\circ}$. The key idea here is to think of 75° as half of some other angle. What angle, when divided by 2, gives us 75°? That's right, it's 150°! So, we can rewrite our problem as finding $\sin(\frac{150^{\circ}}{2})$.

Now, we can directly apply the half-angle formula for sine:

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

In our case, $\theta = 150^{\circ}$. So, we need to find $\cos 150^{\circ}$. Remember, 150° is in the second quadrant, where cosine is negative. It's also related to the 30° reference angle (180° - 150° = 30°). We know that $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, so $\cos 150^{\circ} = -\frac{\sqrt{3}}{2}$. This step is critical because the correct sign of the cosine value will directly impact the final result for $\sin 75^{\circ}$.

Before we plug everything into the formula, let's pause for a moment to consider the sign of $\sin 75^{\circ}$ itself. The angle 75° lies in the first quadrant, where sine is positive. This means we'll choose the positive square root when we apply the half-angle formula. Keeping track of the quadrant is essential to ensure we select the appropriate sign for the trigonometric function, as using the wrong sign would lead to an incorrect answer. This preliminary analysis helps us to avoid errors and reinforces our understanding of the relationship between angles and their trigonometric values.

This step-by-step approach ensures clarity and minimizes the chances of making mistakes. By first identifying the appropriate angle to use in the half-angle formula, then determining the cosine value of that angle, and finally considering the sign of the sine function in the relevant quadrant, we lay a solid foundation for applying the formula correctly and arriving at the accurate solution. This methodical approach is a valuable skill in trigonometry and beyond, as it promotes careful thinking and problem-solving abilities.

Applying the Half-Angle Formula

Alright, we've got all the pieces we need! Let's plug $\cos 150^{\circ} = -\frac{\sqrt{3}}{2}$ into the half-angle formula for sine:

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

Remember, we're using the positive square root because 75° is in the first quadrant. Now, substitute the value of $\cos 150^{\circ}$:

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

See? It's all coming together! The next step is to simplify the expression inside the square root. This involves a bit of algebraic manipulation, but don't worry, we'll take it one step at a time. First, let's get rid of the double negative:

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

Now, to make things easier to work with, let's find a common denominator for the terms in the numerator. We can rewrite 1 as $\frac{2}{2}$:

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

Combine the fractions in the numerator:

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

To divide by 2, we can multiply by its reciprocal, which is $\frac{1}{2}$:

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

Multiply the fractions:

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

We're almost there! Now we have a single fraction under the square root, which makes it easier to simplify. The next step involves taking the square root of both the numerator and the denominator separately.

Simplifying the Expression

We've reached a crucial point where we need to simplify the expression $\sqrt{\frac{2 + \sqrt{3}}{4}}$. Remember, the square root of a fraction is the fraction of the square roots, so we can rewrite this as:

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

The square root of 4 is simply 2, so we have:

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

Now, this might look like our final answer, but we can actually simplify the numerator further. This involves a little trickery with nested square roots. We want to see if we can rewrite $\sqrt{2 + \sqrt{3}}$ in the form $a + b$, where a and b are simpler expressions, possibly involving square roots themselves. To do this, we can try to express $2 + \sqrt{3}$ as a perfect square. This is where our algebraic skills come into play.

To simplify this nested square root, we can use a clever technique. Let's assume that $\sqrt{2 + \sqrt{3}}$ can be written in the form $\sqrt{a} + \sqrt{b}$. Squaring both sides of this equation gives us:

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

Now, we can equate the rational and irrational parts. This gives us two equations:

• $a + b = 2$
• $2\sqrt{ab} = \sqrt{3}$, which simplifies to $4ab = 3$

We now have a system of two equations with two variables. We can solve for a and b. One way to do this is to solve the first equation for one variable (say, $b = 2 - a$) and substitute it into the second equation:

$4a(2 - a) = 3$

Expanding and rearranging, we get a quadratic equation:

$4a^2 - 8a + 3 = 0$

We can solve this quadratic equation using the quadratic formula or by factoring. Factoring gives us:

$(2a - 1)(2a - 3) = 0$

So, the solutions for a are $\frac{1}{2}$ and $\frac{3}{2}$. If $a = \frac{1}{2}$, then $b = 2 - \frac{1}{2} = \frac{3}{2}$. If $a = \frac{3}{2}$, then $b = 2 - \frac{3}{2} = \frac{1}{2}$. Either way, we have the same pair of values for a and b.

Therefore, we can write $\sqrt{2 + \sqrt{3}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}$.

The Final Answer

Okay, let's put it all together! We found that $\sqrt{2 + \sqrt{3}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}$. We can rewrite this as:

$\sqrt{2 + \sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}}$

To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{2}$:

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

Now, substitute this back into our expression for $\sin 75^{\circ}$:

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

Simplify by multiplying the denominator:

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

And there you have it! We've successfully found the value of $\sin 75^{\circ}$ using the half-angle formula. The final answer is:

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

Woohoo! Give yourself a pat on the back, guys. That was a bit of a journey, but we made it through! We started with the half-angle formula, identified the correct angle to use, carefully applied the formula, simplified the expression, and even tackled a nested square root. This problem is a great example of how trigonometry can involve a blend of different algebraic techniques, and it's a testament to the power of these formulas in solving seemingly complex problems.

Conclusion

So, there you have it! We've successfully navigated the twists and turns of trigonometry to find that $\sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$ using the half-angle formula. This journey demonstrates the power and versatility of trigonometric identities, allowing us to break down complex problems into manageable steps. Remember, the key to mastering these concepts is practice, so keep exploring and applying these techniques. Whether you're a student tackling homework or a math enthusiast diving deeper into the subject, the half-angle formulas are a valuable tool in your mathematical arsenal. Keep up the great work, and happy calculating!