Question
\( \cos \hat{A} \cos \hat{B}+\sin \hat{A} \sin \hat{B} \) if \( \hat{X}=225^{\circ} \) and \( \hat{B}=135^{\circ} \)
Ask by Burton Mullins. in South Africa
Mar 10,2025
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Answer
The expression \( \cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B} \) simplifies to 0 when \( \hat{A} = 225^\circ \) and \( \hat{B} = 135^\circ \).
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To solve \( \cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B} \), we can use the cosine addition formula: \[ \cos(\hat{A} - \hat{B}) = \cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B} \] Given \( \hat{A} = 225^\circ \) and \( \hat{B} = 135^\circ \), we find: \[ \hat{A} - \hat{B} = 225^\circ - 135^\circ = 90^\circ \] Now, we compute: \[ \cos(90^\circ) = 0 \] Thus, \( \cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B} = 0 \).
