मराठी

Sin 47° + Sin 61° − Sin 11° − Sin 25° is Equal to

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प्रश्न

sin 47° + sin 61° − sin 11° − sin 25° is equal to

पर्याय

  • sin 36°

  • cos 36°

  • sin 7°

  • cos 7°

MCQ
बेरीज
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उत्तर

cos 7°
\[\sin47^\circ + \sin61^\circ - \sin11^\circ - \sin25^\circ\]
\[ = \sin47^\circ - \sin25^\circ + \sin61^\circ - \sin11^\circ\]
\[ = 2\sin\left( \frac{47^\circ - 25^\circ}{2} \right)\cos\left( \frac{47^\circ + 25^\circ}{2} \right) + 2\sin\left( \frac{61^\circ - 11^\circ}{2} \right)\cos\left( \frac{61^\circ + 11^\circ}{2} \right)\]
\[ = 2\sin11^\circ\cos36^\circ + 2\sin25^\circ\cos36^\circ\]
\[ = 2\cos36^\circ\left( \sin11^\circ + \sin25^\circ \right)\]
\[ = 2\cos36^\circ\left\{ 2\sin\left( \frac{11^\circ + 25^\circ}{2} \right)\cos\left( \frac{11^\circ - 25^\circ}{2} \right) \right\}\]
\[ = 4\cos36^\circ\sin18^\circ\cos7^\circ\]
\[ = 4 \times \left( \frac{\sqrt{5} - 1}{4} \right)\left( \frac{\sqrt{5} + 1}{4} \right)\cos7^\circ \left[ \cos36^\circ = \frac{\sqrt{5} + 1}{4}\text{ and }\sin18^\circ = \frac{\sqrt{5} - 1}{4} \right]\]
\[ = \frac{5 - 1}{4}\cos7^\circ\]
\[ = \cos7^\circ\]

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Transformation Formulae
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Transformation formulae - Exercise 8.4 [पृष्ठ २१]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.4 | Q 9 | पृष्ठ २१

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