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

Numerical

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

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Solution

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

$LHS = \sin^2 72° - \sin^2 6°$

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

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

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

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

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

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 1 | Page 42
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