Find The Exact Value Of Cos(c+d)

Ever found yourself staring at a trigonometry problem, wondering how to break down a complex expression like $\cos(c+d)$ into something manageable? Well, you're in the right place! Today, we're diving deep into a problem that involves finding the exact value of $\cos(c+d)$, given specific information about the sine of angle $c$ and the cosine of angle $d$, along with their respective quadrants. This isn't just about crunching numbers; it's about understanding the fundamental relationships in trigonometry and how angles in different quadrants affect their sine and cosine values. We'll be using the angle addition formula for cosine, which states that $\cos(c+d) = \cos c \cos d - \sin c \sin d$. To apply this formula, we need the values of $\cos c$, $\cos d$, $\sin c$, and $\sin d$. We're given $\sin c = \frac{24}{25}$ and $\cos d = -\frac{3}{4}$. The crucial pieces of information that are missing are $\cos c$ and $\sin d$. Furthermore, we are told that $c$ is in Quadrant II and $d$ is in Quadrant III. These quadrant details are super important because they dictate the signs of our trigonometric functions. Let's get started on unraveling this! Our goal is to meticulously determine each of these values, ensuring accuracy based on the given quadrant information, and then substitute them into the angle addition formula to arrive at our final, exact answer. This process will not only solve the problem at hand but also reinforce your understanding of trigonometric identities and the unit circle.

Understanding the Given Information and the Goal

So, what exactly are we working with? We're given $\sin c = \frac{24}{25}$ and that angle $c$ lies in Quadrant II. Remember, in Quadrant II, the x-coordinates are negative and the y-coordinates are positive. Since sine is related to the y-coordinate on the unit circle, a positive sine value is consistent with Quadrant II. Our first task here is to find the value of $\cos c$. We can use the Pythagorean identity: $\sin^2 c + \cos^2 c = 1$. Substituting the given value of $\sin c$, we get $(\frac{24}{25})^2 + \cos^2 c = 1$. This simplifies to $\frac{576}{625} + \cos^2 c = 1$. To find $\cos^2 c$, we subtract $\frac{576}{625}$ from both sides: $\cos^2 c = 1 - \frac{576}{625} = \frac{625 - 576}{625} = \frac{49}{625}$. Now, taking the square root of both sides, we get $\cos c = \pm \sqrt{\frac{49}{625}} = \pm \frac{7}{25}$. Since angle $c$ is in Quadrant II, where the cosine (x-coordinate) is negative, we must choose the negative value. Therefore, $\cos c = -\frac{7}{25}$.

Now, let's shift our focus to angle $d$. We are given $\cos d = -\frac{3}{4}$ and that $d$ is in Quadrant III. In Quadrant III, both the x-coordinate and the y-coordinate are negative. A negative cosine value is consistent with Quadrant III. Our next step is to find the value of $\sin d$. Again, we'll use the Pythagorean identity: $\sin^2 d + \cos^2 d = 1$. Substituting the given value of $\cos d$, we have $\sin^2 d + (-\frac{3}{4})^2 = 1$. This becomes $\sin^2 d + \frac{9}{16} = 1$. Subtracting $\frac{9}{16}$ from both sides gives us $\sin^2 d = 1 - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16}$. Taking the square root of both sides, we get $\sin d = \pm \sqrt{\frac{7}{16}} = \pm \frac{\sqrt{7}}{4}$. Since angle $d$ is in Quadrant III, where the sine (y-coordinate) is negative, we must choose the negative value. Therefore, $\sin d = -\frac{\sqrt{7}}{4}$.

With all the necessary components in place – $\sin c = \frac{24}{25}$, $\cos c = -\frac{7}{25}$, $\cos d = -\frac{3}{4}$, and $\sin d = -\frac{\sqrt{7}}{4}$ – we are now perfectly positioned to apply the angle addition formula for cosine and find the exact value of $\cos(c+d)$. This methodical approach, breaking down the problem into finding individual trigonometric values based on quadrant information, is key to successfully navigating these types of problems.

Applying the Angle Addition Formula for Cosine

Now that we have all the pieces of the puzzle, it's time to put them together using the angle addition formula for cosine: $\cos(c+d) = \cos c \cos d - \sin c \sin d$. We have painstakingly determined the following values:

• $\sin c = \frac{24}{25}$
• $\cos c = -\frac{7}{25}$
• $\cos d = -\frac{3}{4}$
• $\sin d = -\frac{\sqrt{7}}{4}$

Let's substitute these values into the formula:

$\cos(c+d) = \left(-\frac{7}{25}\right) \left(-\frac{3}{4}\right) - \left(\frac{24}{25}\right) \left(-\frac{\sqrt{7}}{4}\right)$

First, let's calculate the product of the cosines: $\cos c \cos d = (-\frac{7}{25})(-\frac{3}{4})$. When multiplying two negative numbers, the result is positive. So, $(-\frac{7}{25})(-\frac{3}{4}) = \frac{(-7) \times (-3)}{25 \times 4} = \frac{21}{100}$.

Next, let's calculate the product of the sines: $\sin c \sin d = (\frac{24}{25})(-\frac{\sqrt{7}}{4})$. Here, we are multiplying a positive number by a negative number, so the result will be negative. So, $(\frac{24}{25})(-\frac{\sqrt{7}}{4}) = \frac{24 \times (-\sqrt{7})}{25 \times 4} = \frac{-24\sqrt{7}}{100}$.

Now, we substitute these products back into the main formula: $\cos(c+d) = \frac{21}{100} - \left(\frac{-24\sqrt{7}}{100}\right)$.

Subtracting a negative number is the same as adding its positive counterpart. So, $\cos(c+d) = \frac{21}{100} + \frac{24\sqrt{7}}{100}$.

Since both terms have the same denominator, we can combine them into a single fraction: $\cos(c+d) = \frac{21 + 24\sqrt{7}}{100}$.

This is our exact value for $\cos(c+d)$. We've successfully navigated the problem by carefully determining each required trigonometric value, paying close attention to the quadrant information, and applying the angle addition formula correctly. This step-by-step process ensures that our final answer is not only accurate but also derived from a solid understanding of trigonometric principles. It's a great feeling when all the steps align to produce a clean, exact result!

Verifying Quadrant Information and Trigonometric Signs

Let's take a moment to really solidify why understanding the quadrants is so critical in problems like these. The unit circle is our best friend here. Imagine a circle with a radius of 1 centered at the origin. Any point on this circle can be represented by coordinates $(x, y)$. For any angle $\theta$ measured from the positive x-axis, the cosine of that angle, $\cos \theta$, corresponds to the x-coordinate of the point, and the sine of that angle, $\sin \theta$, corresponds to the y-coordinate.

• Quadrant I (0° to 90° or 0 to $\pi/2$ radians): Both x and y are positive. So, $\cos \theta > 0$ and $\sin \theta > 0$. All trigonometric functions are positive here.
• Quadrant II (90° to 180° or $\pi/2$ to $\pi$ radians): x is negative, and y is positive. So, $\cos \theta < 0$ and $\sin \theta > 0$. Only sine (and its reciprocal, cosecant) is positive here.
• Quadrant III (180° to 270° or $\pi$ to $3\pi/2$ radians): Both x and y are negative. So, $\cos \theta < 0$ and $\sin \theta < 0$. Only tangent (and its reciprocal, cotangent) is positive here.
• Quadrant IV (270° to 360° or $3\pi/2$ to $2\pi$ radians): x is positive, and y is negative. So, $\cos \theta > 0$ and $\sin \theta < 0$. Only cosine (and its reciprocal, secant) is positive here.

In our problem, we were given that $c$ is in Quadrant II. This immediately tells us that $\cos c$ must be negative and $\sin c$ must be positive. We were given $\sin c = \frac{24}{25}$, which is positive, fitting perfectly with Quadrant II. When we used the Pythagorean identity $\sin^2 c + \cos^2 c = 1$ and found $\cos^2 c = \frac{49}{625}$, we had two possible values for $\cos c$: $\frac{7}{25}$ and $-\frac{7}{25}$. Because $c$ is in Quadrant II, we had to select $\cos c = -\frac{7}{25}$. If we had mistakenly chosen the positive value, our final answer for $\cos(c+d)$ would have been incorrect.

Similarly, we were given that $d$ is in Quadrant III. This means both $\cos d$ and $\sin d$ must be negative. We were given $\cos d = -\frac{3}{4}$, which is negative, fitting Quadrant III. When we used $\sin^2 d + \cos^2 d = 1$ and found $\sin^2 d = \frac{7}{16}$, we had two possibilities for $\sin d$: $\frac{\sqrt{7}}{4}$ and $-\frac{\sqrt{7}}{4}$. Since $d$ is in Quadrant III, we had to select $\sin d = -\frac{\sqrt{7}}{4}$. Choosing the positive value would have led to an incorrect result.

This careful consideration of quadrant information ensures that we use the correct signs for our trigonometric functions. The Pythagorean identity gives us the magnitude of the trigonometric ratios, but the quadrant tells us the sign. Without this step, the entire calculation would be flawed. It’s a fundamental aspect of trigonometry that cannot be overlooked, especially when dealing with inverse trigonometric functions or solving equations where multiple angles might satisfy a given condition.

Final Check and Conclusion

We've successfully found the exact value of $\cos(c+d)$ to be $\frac{21 + 24\sqrt{7}}{100}$. Let's quickly recap the steps to ensure everything is sound. We identified the need for the angle addition formula $\cos(c+d) = \cos c \cos d - \sin c \sin d$. We were given $\sin c = \frac{24}{25}$ (c in QII) and $\cos d = -\frac{3}{4}$ (d in QIII). Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ and the quadrant information, we deduced:

• For $c$ in Quadrant II: $\cos c = -\frac{7}{25}$ (since sine is positive, cosine must be negative).
• For $d$ in Quadrant III: $\sin d = -\frac{\sqrt{7}}{4}$ (since cosine is negative, sine must also be negative).

Substituting these values into the formula:

$\cos(c+d) = \left(-\frac{7}{25}\right) \left(-\frac{3}{4}\right) - \left(\frac{24}{25}\right) \left(-\frac{\sqrt{7}}{4}\right)$

$\cos(c+d) = \frac{21}{100} - \left(-\frac{24\sqrt{7}}{100}\right)$

$\cos(c+d) = \frac{21}{100} + \frac{24\sqrt{7}}{100}$

$\cos(c+d) = \frac{21 + 24\sqrt{7}}{100}$

The calculation appears to be correct. The result is an exact value, expressed in terms of integers and a square root, which is typical for these kinds of problems. The use of the angle addition formula and careful application of quadrant rules were essential. This problem highlights the interconnectedness of trigonometric identities and the importance of visualizing angles on the unit circle.

For further exploration into trigonometric identities and the unit circle, you can visit the Khan Academy Mathematics section. They offer excellent resources and practice problems that can deepen your understanding.