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To find the exact value of [tex]\(\cos (c + d)\)[/tex], we can use the provided information and the angle addition formula for cosine. Here's the step-by-step solution:
Given:
- [tex]\(\sin c = \frac{24}{25}\)[/tex] where [tex]\(c\)[/tex] is in Quadrant II
- [tex]\(\cos d = -\frac{3}{4}\)[/tex] where [tex]\(d\)[/tex] is in Quadrant III
### Step 1: Determine [tex]\(\cos c\)[/tex]
Since [tex]\(c\)[/tex] is in Quadrant II, [tex]\(\cos c\)[/tex] will be negative. We use the Pythagorean identity:
[tex]\[ \sin^2(c) + \cos^2(c) = 1 \][/tex]
Given [tex]\(\sin c = \frac{24}{25}\)[/tex]:
[tex]\[ \left(\frac{24}{25}\right)^2 + \cos^2(c) = 1 \][/tex]
[tex]\[ \frac{576}{625} + \cos^2(c) = 1 \][/tex]
Solving for [tex]\(\cos^2(c)\)[/tex]:
[tex]\[ \cos^2(c) = 1 - \frac{576}{625} = \frac{625}{625} - \frac{576}{625} = \frac{49}{625} \][/tex]
Thus:
[tex]\[ \cos(c) = -\sqrt{\frac{49}{625}} = -\frac{7}{25} \][/tex]
### Step 2: Determine [tex]\(\sin d\)[/tex]
Since [tex]\(d\)[/tex] is in Quadrant III, [tex]\(\sin d\)[/tex] will be negative. Again, we use the Pythagorean identity:
[tex]\[ \sin^2(d) + \cos^2(d) = 1 \][/tex]
Given [tex]\(\cos d = -\frac{3}{4}\)[/tex]:
[tex]\[ \sin^2(d) + \left(-\frac{3}{4}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(d) + \frac{9}{16} = 1 \][/tex]
Solving for [tex]\(\sin^2(d)\)[/tex]:
[tex]\[ \sin^2(d) = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \][/tex]
Thus:
[tex]\[ \sin(d) = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4} \][/tex]
### Step 3: Use the angle addition formula
The cosine of the sum of two angles is given by:
[tex]\[ \cos(c + d) = \cos c \cos d - \sin c \sin d \][/tex]
Substituting the known values:
[tex]\[ \cos(c + d) = \left(-\frac{7}{25}\right) \left(-\frac{3}{4}\right) - \left(\frac{24}{25}\right) \left(-\frac{\sqrt{7}}{4}\right) \][/tex]
[tex]\[ \cos(c + d) = \frac{21}{100} + \frac{24\sqrt{7}}{100} = \frac{21 + 24\sqrt{7}}{100} \][/tex]
Therefore, the exact value of [tex]\(\cos (c + d)\)[/tex] is:
[tex]\[ \boxed{\frac{21 + 24\sqrt{7}}{100}} \][/tex]
Given:
- [tex]\(\sin c = \frac{24}{25}\)[/tex] where [tex]\(c\)[/tex] is in Quadrant II
- [tex]\(\cos d = -\frac{3}{4}\)[/tex] where [tex]\(d\)[/tex] is in Quadrant III
### Step 1: Determine [tex]\(\cos c\)[/tex]
Since [tex]\(c\)[/tex] is in Quadrant II, [tex]\(\cos c\)[/tex] will be negative. We use the Pythagorean identity:
[tex]\[ \sin^2(c) + \cos^2(c) = 1 \][/tex]
Given [tex]\(\sin c = \frac{24}{25}\)[/tex]:
[tex]\[ \left(\frac{24}{25}\right)^2 + \cos^2(c) = 1 \][/tex]
[tex]\[ \frac{576}{625} + \cos^2(c) = 1 \][/tex]
Solving for [tex]\(\cos^2(c)\)[/tex]:
[tex]\[ \cos^2(c) = 1 - \frac{576}{625} = \frac{625}{625} - \frac{576}{625} = \frac{49}{625} \][/tex]
Thus:
[tex]\[ \cos(c) = -\sqrt{\frac{49}{625}} = -\frac{7}{25} \][/tex]
### Step 2: Determine [tex]\(\sin d\)[/tex]
Since [tex]\(d\)[/tex] is in Quadrant III, [tex]\(\sin d\)[/tex] will be negative. Again, we use the Pythagorean identity:
[tex]\[ \sin^2(d) + \cos^2(d) = 1 \][/tex]
Given [tex]\(\cos d = -\frac{3}{4}\)[/tex]:
[tex]\[ \sin^2(d) + \left(-\frac{3}{4}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(d) + \frac{9}{16} = 1 \][/tex]
Solving for [tex]\(\sin^2(d)\)[/tex]:
[tex]\[ \sin^2(d) = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \][/tex]
Thus:
[tex]\[ \sin(d) = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4} \][/tex]
### Step 3: Use the angle addition formula
The cosine of the sum of two angles is given by:
[tex]\[ \cos(c + d) = \cos c \cos d - \sin c \sin d \][/tex]
Substituting the known values:
[tex]\[ \cos(c + d) = \left(-\frac{7}{25}\right) \left(-\frac{3}{4}\right) - \left(\frac{24}{25}\right) \left(-\frac{\sqrt{7}}{4}\right) \][/tex]
[tex]\[ \cos(c + d) = \frac{21}{100} + \frac{24\sqrt{7}}{100} = \frac{21 + 24\sqrt{7}}{100} \][/tex]
Therefore, the exact value of [tex]\(\cos (c + d)\)[/tex] is:
[tex]\[ \boxed{\frac{21 + 24\sqrt{7}}{100}} \][/tex]
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