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Prove That: Sin 2 2 π 5 − Sin 2 − π 3 = √ 5 − 1 8

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प्रश्न

Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]

  
संख्यात्मक
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उत्तर

\[\frac{2\pi}{5} = 72°, \frac{\pi}{3} = 60° \]

\[LHS = \sin^2 72° - \sin^2 6°\]

\[ = \sin^2 \left( 90° - 18° \right) - \frac{3}{4}\]

\[ = \cos^2 18 °- \frac{3}{4} \left( \because \sin\left( 90°  - \theta \right) = cos\theta \right)\]

\[ = \left( \frac{\sqrt{10 + 2\sqrt{5}}}{4} \right)^2 - \frac{3}{4} \left( \because \cos18°  = \frac{\sqrt{10 + 2\sqrt{5}}}{4} \right)\]

\[ = \frac{10 + 2\sqrt{5}}{16} - \frac{3}{4}\]

\[= \frac{10 + 2\sqrt{5} - 12}{16}\]
\[ = \frac{2\sqrt{5} - 2}{16}\]
\[ = \frac{\sqrt{5} - 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]

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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 1 | पृष्ठ ४२

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