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To find the value of [tex]\(\cos(A + B)\)[/tex], we need to first determine [tex]\(\sin A\)[/tex], [tex]\(\cos A\)[/tex], and [tex]\(\sin B\)[/tex] from the given information.
Given:
[tex]\[ \tan A = \frac{35}{12} \quad \text{and} \quad \cos B = \frac{28}{53} \][/tex]
Step 1: Determine [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex].
Using the trigonometric identity for [tex]\(\tan A\)[/tex]:
[tex]\[ \tan A = \frac{\sin A}{\cos A} \][/tex]
We have:
[tex]\[ \tan A = \frac{35}{12} \][/tex]
This can be represented as:
[tex]\[ \sin A = 35k \quad \text{and} \quad \cos A = 12k \][/tex]
We use the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Substitute the expressions for [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
[tex]\[ (35k)^2 + (12k)^2 = 1 \][/tex]
[tex]\[ 1225k^2 + 144k^2 = 1 \][/tex]
[tex]\[ 1369k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{1369} \][/tex]
[tex]\[ k = \frac{1}{\sqrt{1369}} = \frac{1}{37} \][/tex]
Now, find [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
[tex]\[ \sin A = 35k = 35 \times \frac{1}{37} = \frac{35}{37} \][/tex]
[tex]\[ \cos A = 12k = 12 \times \frac{1}{37} = \frac{12}{37} \][/tex]
Step 2: Determine [tex]\(\sin B\)[/tex].
Use the Pythagorean identity:
[tex]\[ \sin^2 B + \cos^2 B = 1 \][/tex]
Given [tex]\(\cos B\)[/tex]:
[tex]\[ \cos B = \frac{28}{53} \][/tex]
[tex]\[ \cos^2 B = \left(\frac{28}{53}\right)^2 = \frac{784}{2809} \][/tex]
Thus:
[tex]\[ \sin^2 B + \frac{784}{2809} = 1 \][/tex]
[tex]\[ \sin^2 B = 1 - \frac{784}{2809} \][/tex]
[tex]\[ \sin^2 B = \frac{2809}{2809} - \frac{784}{2809} \][/tex]
[tex]\[ \sin^2 B = \frac{2025}{2809} \][/tex]
[tex]\[ \sin B = \sqrt{\frac{2025}{2809}} = \frac{45}{53} \][/tex]
Step 3: Use the angle addition formula for cosine.
[tex]\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \][/tex]
Substitute the known values:
[tex]\[ \cos A = \frac{12}{37}, \quad \cos B = \frac{28}{53} \][/tex]
[tex]\[ \sin A = \frac{35}{37}, \quad \sin B = \frac{45}{53} \][/tex]
Calculate:
[tex]\[ \cos(A + B) = \left(\frac{12}{37} \times \frac{28}{53}\right) - \left(\frac{35}{37} \times \frac{45}{53}\right) \][/tex]
First term:
[tex]\[ \frac{12 \times 28}{37 \times 53} = \frac{336}{1961} \][/tex]
Second term:
[tex]\[ \frac{35 \times 45}{37 \times 53} = \frac{1575}{1961} \][/tex]
So:
[tex]\[ \cos(A + B) = \frac{336}{1961} - \frac{1575}{1961} = \frac{336 - 1575}{1961} = \frac{-1239}{1961} \][/tex]
Thus, the value of [tex]\(\cos(A + B)\)[/tex] is:
[tex]\[ \cos(A + B) = \boxed{-\frac{1239}{1961}} \][/tex]
Given:
[tex]\[ \tan A = \frac{35}{12} \quad \text{and} \quad \cos B = \frac{28}{53} \][/tex]
Step 1: Determine [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex].
Using the trigonometric identity for [tex]\(\tan A\)[/tex]:
[tex]\[ \tan A = \frac{\sin A}{\cos A} \][/tex]
We have:
[tex]\[ \tan A = \frac{35}{12} \][/tex]
This can be represented as:
[tex]\[ \sin A = 35k \quad \text{and} \quad \cos A = 12k \][/tex]
We use the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Substitute the expressions for [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
[tex]\[ (35k)^2 + (12k)^2 = 1 \][/tex]
[tex]\[ 1225k^2 + 144k^2 = 1 \][/tex]
[tex]\[ 1369k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{1369} \][/tex]
[tex]\[ k = \frac{1}{\sqrt{1369}} = \frac{1}{37} \][/tex]
Now, find [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]:
[tex]\[ \sin A = 35k = 35 \times \frac{1}{37} = \frac{35}{37} \][/tex]
[tex]\[ \cos A = 12k = 12 \times \frac{1}{37} = \frac{12}{37} \][/tex]
Step 2: Determine [tex]\(\sin B\)[/tex].
Use the Pythagorean identity:
[tex]\[ \sin^2 B + \cos^2 B = 1 \][/tex]
Given [tex]\(\cos B\)[/tex]:
[tex]\[ \cos B = \frac{28}{53} \][/tex]
[tex]\[ \cos^2 B = \left(\frac{28}{53}\right)^2 = \frac{784}{2809} \][/tex]
Thus:
[tex]\[ \sin^2 B + \frac{784}{2809} = 1 \][/tex]
[tex]\[ \sin^2 B = 1 - \frac{784}{2809} \][/tex]
[tex]\[ \sin^2 B = \frac{2809}{2809} - \frac{784}{2809} \][/tex]
[tex]\[ \sin^2 B = \frac{2025}{2809} \][/tex]
[tex]\[ \sin B = \sqrt{\frac{2025}{2809}} = \frac{45}{53} \][/tex]
Step 3: Use the angle addition formula for cosine.
[tex]\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \][/tex]
Substitute the known values:
[tex]\[ \cos A = \frac{12}{37}, \quad \cos B = \frac{28}{53} \][/tex]
[tex]\[ \sin A = \frac{35}{37}, \quad \sin B = \frac{45}{53} \][/tex]
Calculate:
[tex]\[ \cos(A + B) = \left(\frac{12}{37} \times \frac{28}{53}\right) - \left(\frac{35}{37} \times \frac{45}{53}\right) \][/tex]
First term:
[tex]\[ \frac{12 \times 28}{37 \times 53} = \frac{336}{1961} \][/tex]
Second term:
[tex]\[ \frac{35 \times 45}{37 \times 53} = \frac{1575}{1961} \][/tex]
So:
[tex]\[ \cos(A + B) = \frac{336}{1961} - \frac{1575}{1961} = \frac{336 - 1575}{1961} = \frac{-1239}{1961} \][/tex]
Thus, the value of [tex]\(\cos(A + B)\)[/tex] is:
[tex]\[ \cos(A + B) = \boxed{-\frac{1239}{1961}} \][/tex]
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