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प्रश्न
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
पर्याय
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0
-1
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उत्तर
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\[\text{ We have,} \]
\[2\text{ cos } x - \cos3x - \cos5x - 16 \cos^3 x \sin^2 x\]
\[ = 2\text{ cos } x - \cos3x - \cos5x - 16\left[ \frac{\cos3x + 3\text{ cos } x}{4} \times \frac{\left( 1 - \cos2x \right)}{2} \right]\]
\[ = 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \left( \cos3x + 3\text{ cos } x \right)\left( 1 - \cos2x \right) \right]\]
\[ = 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \cos3x - \cos3x \cos2x + 3\text{ cos } x - 3\text{ cos } x \cos2x \right]\]
\[ = 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \cos3x + 3\text{ cos } x \right] + 2\cos3x \cos2x + 3\left[ 2\text{ cos } x \cos2x \right]\]
\[ = 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \cos3x + 3\text{ cos } x \right] + \cos5x + \text{ cos } x + 3\cos3x + 3\text{ cos } x\]
\[ \left[ 2cosAcosB = \cos\left( A + B \right) + \cos\left( A - B \right) \right]\]
\[ = 2\text{ cos } x - \cos3x - \cos5x - 2\cos3x - 6\text{ cos } x + \cos5x + \text{ cos } x + 3\cos3x + 3\text{ cos } x = 0\]
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