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To prove that if [tex]\(\operatorname{Sin} \alpha + \operatorname{Sin} \beta = \frac{1}{4}\)[/tex] and [tex]\(\operatorname{Cos} \alpha + \operatorname{Cos} \beta = \frac{1}{2}\)[/tex], then [tex]\(\tan \left( \frac{\alpha + \beta}{2} \right) = \frac{1}{2}\)[/tex], follow these steps:
1. Set up expressions using sum-to-product identities:
Let [tex]\(A = \frac{\alpha + \beta}{2}\)[/tex] and [tex]\(B = \frac{\alpha - \beta}{2}\)[/tex]. Then we can use the sum-to-product identities:
[tex]\[ \operatorname{Sin} \alpha + \operatorname{Sin} \beta = 2 \operatorname{Sin}\left( \frac{\alpha + \beta}{2} \right) \operatorname{Cos}\left( \frac{\alpha - \beta}{2} \right) = 2 \operatorname{Sin}(A) \operatorname{Cos}(B) \][/tex]
[tex]\[ \operatorname{Cos} \alpha + \operatorname{Cos} \beta = 2 \operatorname{Cos}\left( \frac{\alpha + \beta}{2} \right) \operatorname{Cos}\left( \frac{\alpha - \beta}{2} \right) = 2 \operatorname{Cos}(A) \operatorname{Cos}(B) \][/tex]
2. Substitute the given values:
Given [tex]\(\operatorname{Sin} \alpha + \operatorname{Sin} \beta = \frac{1}{4}\)[/tex]:
[tex]\[ 2 \operatorname{Sin}(A) \operatorname{Cos}(B) = \frac{1}{4} \][/tex]
[tex]\[ \operatorname{Sin}(A) \operatorname{Cos}(B) = \frac{1}{8} \quad \text{(1)} \][/tex]
Given [tex]\(\operatorname{Cos} \alpha + \operatorname{Cos} \beta = \frac{1}{2}\)[/tex]:
[tex]\[ 2 \operatorname{Cos}(A) \operatorname{Cos}(B) = \frac{1}{2} \][/tex]
[tex]\[ \operatorname{Cos}(A) \operatorname{Cos}(B) = \frac{1}{4} \quad \text{(2)} \][/tex]
3. Divide equation (1) by equation (2):
[tex]\[ \frac{\operatorname{Sin}(A) \operatorname{Cos}(B)}{\operatorname{Cos}(A) \operatorname{Cos}(B)} = \frac{\frac{1}{8}}{\frac{1}{4}} \][/tex]
[tex]\[ \frac{\operatorname{Sin}(A)}{\operatorname{Cos}(A)} = \frac{\frac{1}{8}}{\frac{1}{4}} \][/tex]
[tex]\[ \frac{\operatorname{Sin}(A)}{\operatorname{Cos}(A)} = \frac{1}{2} \][/tex]
4. Simplify to find [tex]\(\tan(A)\)[/tex]:
[tex]\[ \operatorname{Tan}(A) = \frac{1}{2} \][/tex]
Thus, we have shown that:
[tex]\[ \tan \left( \frac{\alpha + \beta}{2} \right) = \frac{1}{2} \][/tex]
1. Set up expressions using sum-to-product identities:
Let [tex]\(A = \frac{\alpha + \beta}{2}\)[/tex] and [tex]\(B = \frac{\alpha - \beta}{2}\)[/tex]. Then we can use the sum-to-product identities:
[tex]\[ \operatorname{Sin} \alpha + \operatorname{Sin} \beta = 2 \operatorname{Sin}\left( \frac{\alpha + \beta}{2} \right) \operatorname{Cos}\left( \frac{\alpha - \beta}{2} \right) = 2 \operatorname{Sin}(A) \operatorname{Cos}(B) \][/tex]
[tex]\[ \operatorname{Cos} \alpha + \operatorname{Cos} \beta = 2 \operatorname{Cos}\left( \frac{\alpha + \beta}{2} \right) \operatorname{Cos}\left( \frac{\alpha - \beta}{2} \right) = 2 \operatorname{Cos}(A) \operatorname{Cos}(B) \][/tex]
2. Substitute the given values:
Given [tex]\(\operatorname{Sin} \alpha + \operatorname{Sin} \beta = \frac{1}{4}\)[/tex]:
[tex]\[ 2 \operatorname{Sin}(A) \operatorname{Cos}(B) = \frac{1}{4} \][/tex]
[tex]\[ \operatorname{Sin}(A) \operatorname{Cos}(B) = \frac{1}{8} \quad \text{(1)} \][/tex]
Given [tex]\(\operatorname{Cos} \alpha + \operatorname{Cos} \beta = \frac{1}{2}\)[/tex]:
[tex]\[ 2 \operatorname{Cos}(A) \operatorname{Cos}(B) = \frac{1}{2} \][/tex]
[tex]\[ \operatorname{Cos}(A) \operatorname{Cos}(B) = \frac{1}{4} \quad \text{(2)} \][/tex]
3. Divide equation (1) by equation (2):
[tex]\[ \frac{\operatorname{Sin}(A) \operatorname{Cos}(B)}{\operatorname{Cos}(A) \operatorname{Cos}(B)} = \frac{\frac{1}{8}}{\frac{1}{4}} \][/tex]
[tex]\[ \frac{\operatorname{Sin}(A)}{\operatorname{Cos}(A)} = \frac{\frac{1}{8}}{\frac{1}{4}} \][/tex]
[tex]\[ \frac{\operatorname{Sin}(A)}{\operatorname{Cos}(A)} = \frac{1}{2} \][/tex]
4. Simplify to find [tex]\(\tan(A)\)[/tex]:
[tex]\[ \operatorname{Tan}(A) = \frac{1}{2} \][/tex]
Thus, we have shown that:
[tex]\[ \tan \left( \frac{\alpha + \beta}{2} \right) = \frac{1}{2} \][/tex]
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