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Prove that: sin 2 42 ° − cos 2 78 = √ 5 + 1 8

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प्रश्न

Prove that:  \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\] 

 
संख्यात्मक
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उत्तर

\[LHS = \sin^2 42° - \cos^2 78° \]
\[ = \sin^2 \left( 90°  - 48° \right) - \cos^2 \left( 90° - 12°  \right)\]
\[ = \cos^2 48°  - \sin^2 12°  \]
\[ = \cos\left( 48° + 12°  \right) \cos\left( 48°  - 12°  \right) \left[ \cos\left( A + B \right) \cos\left( A - B \right) = \cos^2 A - \sin^2 \right]\]
\[ = \cos60°  \cos36° \]
\[ = \frac{1}{2} \times \frac{\sqrt{5} + 1}{4} \left( \because \cos36°  = \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{\sqrt{5} + 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]

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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 3 | पृष्ठ ४२

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