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Prove That: Sin ( 3 π 8 − 5 ) Cos ( π 8 + 5 ) + Cos ( 3 π 8 − 5 ) Sin ( π 8 + 5 ) = 1

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प्रश्न

Prove that:

\[\sin\left( \frac{3\pi}{8} - 5 \right)\cos\left( \frac{\pi}{8} + 5 \right) + \cos\left( \frac{3\pi}{8} - 5 \right)\sin\left( \frac{\pi}{8} + 5 \right) = 1\]

 

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उत्तर

LHS =
\[\sin\left( \frac{3\pi}{8} - 5 \right)\cos\left( \frac{\pi}{8} + 5 \right) + \cos\left( \frac{3\pi}{8} - 5 \right)\sin\left( \frac{\pi}{8} + 5 \right)\]
\[ = \sin\left[ \left( \frac{3\pi}{8} - 5 \right) + \left( \frac{\pi}{8} + 5 \right) \right] \left[ \sin A\cos B + \cos A\sin B = \sin\left( A + B \right) \right]\]
\[ = \sin\frac{4\pi}{8}\]
\[ = \sin\frac{\pi}{2}\]
\[ = 1\]

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ १९]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 12.3 | पृष्ठ १९

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