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Prove that Sin 3 X + Sin 2 X − Sin X = 4 Sin X Cos X 2 Cos 3 X 2 - Mathematics

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प्रश्न

Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]

संख्यात्मक
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उत्तर

\[LHS = \sin3x + \sin2x - \text{ sin } x\]

\[ = \sin3x + 2\sin\left( \frac{2x - x}{2} \right)\cos\left( \frac{2x + x}{2} \right) \left[ \because \text{ sin } A - \text{ sin } B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]

\[ = \sin3x + 2\sin\left( \frac{x}{2} \right)\cos\left( \frac{3x}{2} \right)\]

\[ = 2\sin\left( \frac{3x}{2} \right)\cos\left( \frac{3x}{2} \right) + 2\sin\left( \frac{3x}{2} \right)\cos\frac{x}{2} \left[ \because \sin2A = 2\text{ sin } A\text{ cos } A \right]\]

\[ = 2\cos\left( \frac{3x}{2} \right)\left[ \sin\left( \frac{3x}{2} \right)\cos\left( \frac{x}{2} \right) \right]\]

\[ = 2\cos\left( \frac{3x}{2} \right)\left[ 2\sin\left( \frac{\frac{3x}{2} + \frac{x}{2}}{2} \right)\cos\left( \frac{\frac{3x}{2} - \frac{x}{2}}{2} \right) \right] \left[ \because \text{ sin } A + \text{ sin } B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]

\[ = 2\cos\left( \frac{3x}{2} \right)\left[ 2\sin\left( x \right)\cos\left( \frac{x}{2} \right) \right]\]

\[ = 4\sin\left( x \right)\cos\left( \frac{x}{2} \right)\cos\left( \frac{3x}{2} \right) = RHS\]

\[\text{ Hence proved } . \]

 

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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 9.1 [पृष्ठ २८]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 25 | पृष्ठ २८

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