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Trigonometric Ratios of Multiples of an Angle

Prove the fol...

Question

Prove the following:
$\frac{c o s 9 x - c o s 5 x}{s i n 17 x - s i n 3 x} = - \frac{s i n 2 x}{c o s 10 x}$

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Solution

$L H S = \frac{c o s 9 x - c o s 5 x}{s i n 17 x - s i n 3 x} = \frac{- 2 sin (\frac{9 x + 5 x}{2}) sin (\frac{9 x - 5 x}{2})}{2 cos (\frac{17 x + 3 x}{2}) sin (\frac{17 x - 3 x}{2})}$
$= - \frac{sin 7 x s i n 2 x}{c o s 10 x sin 7 x}$

$= - \frac{s i n 2 x}{c o s 10 x} = R H S$
Hence proved

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Trigonometric Ratios of Multiples of an Angle

Standard XI Mathematics

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