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In a ∆Abc, Prove That: Cos ( a + B 2 ) = Sin C 2

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प्रश्न

In a ∆ABC, prove that:

\[\cos\left( \frac{A + B}{2} \right) = \sin\frac{C}{2}\]

 

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उत्तर

 In ∆ ABC: 
\[A + B + C = \pi\]
\[ \Rightarrow A + B = \pi - C\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi}{2} - \frac{C}{2}\]
\[\text{ Now, LHS }= \cos\left( \frac{A + B}{2} \right) \]
\[ = \cos\left( \frac{\pi}{2} - \frac{C}{2} \right) \]
\[ = \sin \left( \frac{C}{2} \right) \left[ \because \cos\left( \frac{\pi}{2} - \theta \right) = \sin \theta \right] \]
 = RHS
Hence proved.

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अध्याय 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ४०]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 5 Trigonometric Functions
Exercise 5.3 | Q 6.2 | पृष्ठ ४०

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