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The Value of Sin 2 5 π 12 − Sin 2 π 12

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प्रश्न

The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\] 

विकल्प

  • (a) \[\frac{1}{2}\] 

     
  • (b) \[\frac{\sqrt{3}}{2}\] 

  • (c) 1 

  • (d) 0 

MCQ
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उत्तर

(b)  \[\frac{\sqrt{3}}{2}\] \[\frac{5\pi}{12} = 75°, \frac{\pi}{12} = 15°\] 

\[\sin^2 75° - \sin^2 15° \]

\[ = \sin^2 75 ° - \cos^2 75° \left[ \sin\left( 90° - \theta \right) = \cos\theta \right]\]

\[\text{ Now }, \sin75° = \sin(45° + 30°)\]

\[ = \sin45°\cos30°+ \cos45°\sin30°\]

\[ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}\]

\[ = \frac{\sqrt{3} + 1}{2\sqrt{2}}\]

\[\cos75°= \cos(45° + 30°)\]

\[ = \cos45° \cos30°- \sin45°\sin30°\]

\[ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}\]

\[ = \frac{\sqrt{3} - 1}{2\sqrt{2}}\]

\[\text{ Hence } , \]

\[ \sin^2 75° - \cos^2 75° = \left( \frac{\sqrt{3} + 1}{2\sqrt{2}} \right)^2 - \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right)^2 \]

\[ = \frac{3 + 1 + 2\sqrt{3} - 3 - 1 + 2\sqrt{3}}{8}\]

\[ = \frac{4\sqrt{3}}{8}\]

\[ = \frac{\sqrt{3}}{2}\]

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.4 [पृष्ठ २७]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.4 | Q 1 | पृष्ठ २७

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