Trigonometry Table: Solving Y = 3 Sin 2θ - Cos Θ

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Hey guys! Today, we're diving deep into the awesome world of trigonometry to tackle a specific problem: completing a table of values for the function $y=3 \sin 2 \theta-\cos \theta$. This might sound a bit daunting at first, but trust me, with a little practice and understanding, you'll be a pro in no time. We're going to break down each step, ensuring you not only get the right answers but also grasp the 'why' behind them. So, grab your calculators, your favorite notebook, and let's get started on this mathematical adventure!

Understanding the Function: $y=3 \sin 2 \theta-\cos \theta$

Before we jump into filling out the table, let's take a moment to appreciate the function we're working with: $y=3 \sin 2 \theta-\cos \theta$. This equation combines two fundamental trigonometric functions: sine and cosine. The '3' in front of the sine term indicates an amplitude change, meaning the sine wave will be stretched vertically. The '$2\theta$' inside the sine function means the wave will be compressed horizontally, completing two cycles for every one cycle of a standard sine wave. Finally, the '$-\cos \theta$' term introduces a cosine function, shifted and possibly reflected. Understanding these individual components is key to predicting the overall behavior of the function. When we plug in different values for $\theta$, we're essentially finding points on the graph of this trigonometric curve. The table we're completing will give us a discrete set of these points, helping us visualize the function's shape and key features.

Step-by-Step Calculation for Each Value

Now, let's get down to business and fill in those blanks in our table. We'll go through each value of $\theta$ systematically, calculating the corresponding 'y' value using the given function $y=3 \sin 2 \theta-\cos \theta$. Remember to set your calculator to degree mode for these calculations!

1. For $\theta = 0^0$:

This one's already given as $y = -1.0$. Let's quickly verify it to make sure we're on the same page.

$y = 3 \sin (2 \times 0^0) - \cos (0^0)$

$y = 3 \sin (0^0) - \cos (0^0)$

We know that $\sin (0^0) = 0$ and $\cos (0^0) = 1$.

$y = 3(0) - 1$

$y = 0 - 1$

$y = -1.0$

Perfect! It matches the given value.

2. For $\theta = 30^0$:

Here, we substitute $\theta = 30^0$ into our function:

$y = 3 \sin (2 \times 30^0) - \cos (30^0)$

$y = 3 \sin (60^0) - \cos (30^0)$

Now, let's recall our special trigonometric values or use our calculator:

$\sin (60^0) = \frac{\sqrt{3}}{2} \approx 0.8660$

$\cos (30^0) = \frac{\sqrt{3}}{2} \approx 0.8660$

Plugging these back in:

$y = 3 \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}$

$y = \frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2}$

$y = \frac{2\sqrt{3}}{2}$

$y = \sqrt{3} \approx 1.732$

So, for $\theta = 30^0$, $y \approx 1.732$.

3. For $\theta = 60^0$:

Let's substitute $\theta = 60^0$:

$y = 3 \sin (2 \times 60^0) - \cos (60^0)$

$y = 3 \sin (120^0) - \cos (60^0)$

Recall or calculate:

$\sin (120^0) = \sin (180^0 - 60^0) = \sin (60^0) = \frac{\sqrt{3}}{2} \approx 0.8660$

$\cos (60^0) = \frac{1}{2} = 0.5$

Substituting these values:

$y = 3 \times \frac{\sqrt{3}}{2} - \frac{1}{2}$

$y = \frac{3\sqrt{3}}{2} - \frac{1}{2}$

$y = \frac{3\sqrt{3} - 1}{2} \approx \frac{3 \times 1.732 - 1}{2} \approx \frac{5.196 - 1}{2} \approx \frac{4.196}{2} \approx 2.098$

So, for $\theta = 60^0$, $y \approx 2.098$.

4. For $\theta = 90^0$:

This value is also given as $y = 0$. Let's check:

$y = 3 \sin (2 \times 90^0) - \cos (90^0)$

$y = 3 \sin (180^0) - \cos (90^0)$

We know $\sin (180^0) = 0$ and $\cos (90^0) = 0$.

$y = 3(0) - 0$

$y = 0$

Confirmed!

5. For $\theta = 120^0$:

Substitute $\theta = 120^0$:

$y = 3 \sin (2 \times 120^0) - \cos (120^0)$

$y = 3 \sin (240^0) - \cos (120^0)$

Let's find these values:

$\sin (240^0) = \sin (180^0 + 60^0) = -\sin (60^0) = -\frac{\sqrt{3}}{2} \approx -0.8660$

$\cos (120^0) = \cos (180^0 - 60^0) = -\cos (60^0) = -\frac{1}{2} = -0.5$

Now, plug them into the equation:

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

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

$y = \frac{1 - 3\sqrt{3}}{2} \approx \frac{1 - 3 \times 1.732}{2} \approx \frac{1 - 5.196}{2} \approx \frac{-4.196}{2} \approx -2.098$

So, for $\theta = 120^0$, $y \approx -2.098$.

6. For $\theta = 150^0$:

Substitute $\theta = 150^0$:

$y = 3 \sin (2 \times 150^0) - \cos (150^0)$

$y = 3 \sin (300^0) - \cos (150^0)$

Find the values:

$\sin (300^0) = \sin (360^0 - 60^0) = -\sin (60^0) = -\frac{\sqrt{3}}{2} \approx -0.8660$

$\cos (150^0) = \cos (180^0 - 30^0) = -\cos (30^0) = -\frac{\sqrt{3}}{2} \approx -0.8660$

Substitute into the function:

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

$y = -\frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2}$

$y = -\frac{2\sqrt{3}}{2}$

$y = -\sqrt{3} \approx -1.732$

So, for $\theta = 150^0$, $y \approx -1.732$.

7. For $\theta = 180^0$:

This last value is also provided as $y = 1.0$. Let's verify:

$y = 3 \sin (2 \times 180^0) - \cos (180^0)$

$y = 3 \sin (360^0) - \cos (180^0)$

We know $\sin (360^0) = 0$ and $\cos (180^0) = -1$.

$y = 3(0) - (-1)$

$y = 0 + 1$

$y = 1.0$

And there we have it, confirmed!

The Completed Table of Values

Alright, guys, after all that hard work, let's put all our calculated values together into the table. This gives us a clear picture of how the function behaves across the given range of angles. Seeing the numbers laid out like this really helps solidify your understanding of the trigonometric relationships at play.

$ \theta \quad | \quad 0^0 \quad | \quad 30^0 \quad | \quad 60^0 \quad | \quad 90^0 \quad | \quad 120^0 \quad | \quad 150^0 \quad | \quad 180^0 \ \hline Y \quad | \quad -1.0 \quad | \quad 1.732 \quad | \quad 2.098 \quad | \quad 0 \quad | \quad -2.098 \quad | \quad -1.732 \quad | \quad 1.0$

Isn't that neat? You've just successfully completed a table of values for a fairly complex trigonometric function. Each of these points $( \theta, y )$ represents a specific location on the graph of $y = 3 \sin 2 \theta - \cos \theta$. If you were to plot these points, you'd start to see the characteristic wave-like shape emerge.

Practical Applications and Further Exploration

So, why do we do this kind of exercise? Completing tables of values for trigonometric functions is a fundamental skill that underpins many areas of mathematics and science. It's not just about crunching numbers; it's about understanding periodic behavior. Think about waves in physics – sound waves, light waves, water waves – they are all modeled using trigonometric functions. In engineering, these functions are crucial for analyzing anything that oscillates or rotates, like AC circuits, signal processing, and mechanical vibrations. Even in computer graphics, trigonometric functions are used to create animations and simulate realistic movements.

The Importance of Accuracy: Throughout these calculations, paying attention to detail and ensuring accuracy is paramount. Small errors in calculating sine or cosine values can snowball, leading to incorrect final results. Always double-check your calculator settings (degrees vs. radians!) and your substitution steps. Understanding the unit circle and the properties of trigonometric functions (like symmetry and periodicity) can also significantly speed up your calculations and improve accuracy, especially for angles beyond the first quadrant. For instance, knowing that $\sin(180^\circ - x) = \sin(x)$ and $\cos(180^\circ - x) = -\cos(x)$ saves you from needing to find the sine and cosine of unfamiliar angles directly.

Visualizing the Graph: Once you have the table of values, the next logical step is to plot these points on a graph. This visual representation makes the abstract numerical data tangible. You can see the peaks, troughs, and where the function crosses the x-axis (the roots or zeros). For $y = 3 \sin 2 \theta - \cos \theta$, you'd observe a complex wave pattern influenced by both the doubled frequency of the sine component and the cosine component. This visualization helps in understanding concepts like amplitude, period, and phase shift more intuitively. Experimenting with graphing tools or software can be incredibly helpful here; you can input the function and see how your calculated points align with the actual curve.

Exploring Other Functions: This exercise is a gateway to exploring a vast array of trigonometric functions. Try changing the coefficients, the multiples of $\theta$, or combining different trigonometric functions. For example, what happens if you try $y = \sin \theta + \cos \theta$, or $y = 2 \tan \theta$? Each variation will produce a unique graph and set of properties. Understanding the impact of each change – how multiplying $\theta$ by 2 affects the period, or how adding a constant shifts the graph vertically – builds a robust understanding of function transformations.

Connecting to Calculus: For those of you further along in your math journey, these trigonometric functions are the building blocks for calculus concepts. Derivatives and integrals of trigonometric functions are fundamental. For instance, the derivative of $\sin \theta$ is $\cos \theta$, and the derivative of $\cos \theta$ is $-\sin \theta$. Understanding the basic functions and their values is the first step to mastering these more advanced topics. The function $y = 3 \sin 2 \theta - \cos \theta$ itself can be differentiated and integrated, leading to more complex calculus problems that still rely on the foundational knowledge gained from completing this table.

Ultimately, mastering these trigonometric calculations isn't just about passing a test; it's about equipping yourself with powerful tools to understand and describe the cyclical and wave-like phenomena that are ubiquitous in our universe. Keep practicing, stay curious, and you'll find that the world of trigonometry is full of fascinating patterns and applications!