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The value of sin 78° − sin 66° − sin 42° + sin 60° is

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प्रश्न

The value of sin 78° − sin 66° − sin 42° + sin 60° is ______.

विकल्प

  • \[\frac{1}{2}\]

     

  • \[- \frac{1}{2}\]

     

  • −1

  • None of these

MCQ
रिक्त स्थान भरें
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उत्तर

None of these

Explanation:

= \[\sin78^\circ - \sin66^\circ - \sin42^\circ + \sin60^\circ\]

\[ = \sin78^\circ - \sin42^\circ - \sin66^\circ + \sin60^\circ\]

\[ = 2\sin\left( \frac{78^\circ - 42^\circ}{2} \right)\cos\left( \frac{78^\circ + 42}{2} \right) - \sin66^\circ + \sin60^\circ \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]

\[ = 2\sin18^\circ \cos60^\circ - \sin66^\circ + \sin60^\circ\]

\[ = 2 \times \frac{1}{2}\sin18^\circ - \sin66^\circ + \frac{\sqrt{3}}{2}\]

\[ = \sin18^\circ - \sin66^\circ + \frac{\sqrt{3}}{2}\]

\[ = \frac{\sqrt{5} - 1}{4} - 0 . 914 + \frac{\sqrt{3}}{2}\]

= 0.309 − 0.914 + 0.866

= 0.261

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Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.4 [पृष्ठ २१]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.4 | Q 5 | पृष्ठ २१

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