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Prove that: sin 2 24 ° − sin 2 6 ° = √ 5 − 1 8

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प्रश्न

Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]

  
संख्यात्मक
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उत्तर

\[LHS = \sin^2 24° - \sin^2 6° \]
\[ = \sin\left( 24°  + 6° \right) \sin\left( 24°  - 6° \right) \left[ \sin\left( A + B \right) \sin\left( A - B \right) = \sin^2 A - \sin^2 B \right]\]
\[ = \sin30° \sin18° \]
\[ = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \left( \because \sin18°  = \frac{\sqrt{5} - 1}{4} \right)\]
\[ = \frac{\sqrt{5} - 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\] 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.3 [पृष्ठ ४२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 2 | पृष्ठ ४२

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