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To express the sum [tex]\(\cos (20.3 t) + \cos (12.3 t)\)[/tex] as a product, we utilize the sum-to-product identities which state:
[tex]\[ \cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \][/tex]
Let's identify [tex]\(A\)[/tex] and [tex]\(B\)[/tex] in our problem:
- [tex]\(A = 20.3 t\)[/tex]
- [tex]\(B = 12.3 t\)[/tex]
Using the sum-to-product identity:
[tex]\[ \cos (20.3 t) + \cos (12.3 t) = 2 \cos \left( \frac{20.3 t + 12.3 t}{2} \right) \cos \left( \frac{20.3 t - 12.3 t}{2} \right) \][/tex]
Let's compute the expressions inside the cosines:
[tex]\[ \frac{20.3 t + 12.3 t}{2} = \frac{32.6 t}{2} = 16.3 t \][/tex]
[tex]\[ \frac{20.3 t - 12.3 t}{2} = \frac{8 t}{2} = 4 t \][/tex]
Substituting these results back into the identity:
[tex]\[ \cos (20.3 t) + \cos (12.3 t) = 2 \cos (16.3 t) \cos (4 t) \][/tex]
Therefore, the sum [tex]\(\cos (20.3 t) + \cos (12.3 t)\)[/tex] can be written as a product:
[tex]\[ \boxed{2 \cos (16.3 t) \cos (4 t)} \][/tex]
This is the final expression that represents the given sum in product form.
[tex]\[ \cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \][/tex]
Let's identify [tex]\(A\)[/tex] and [tex]\(B\)[/tex] in our problem:
- [tex]\(A = 20.3 t\)[/tex]
- [tex]\(B = 12.3 t\)[/tex]
Using the sum-to-product identity:
[tex]\[ \cos (20.3 t) + \cos (12.3 t) = 2 \cos \left( \frac{20.3 t + 12.3 t}{2} \right) \cos \left( \frac{20.3 t - 12.3 t}{2} \right) \][/tex]
Let's compute the expressions inside the cosines:
[tex]\[ \frac{20.3 t + 12.3 t}{2} = \frac{32.6 t}{2} = 16.3 t \][/tex]
[tex]\[ \frac{20.3 t - 12.3 t}{2} = \frac{8 t}{2} = 4 t \][/tex]
Substituting these results back into the identity:
[tex]\[ \cos (20.3 t) + \cos (12.3 t) = 2 \cos (16.3 t) \cos (4 t) \][/tex]
Therefore, the sum [tex]\(\cos (20.3 t) + \cos (12.3 t)\)[/tex] can be written as a product:
[tex]\[ \boxed{2 \cos (16.3 t) \cos (4 t)} \][/tex]
This is the final expression that represents the given sum in product form.
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