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Prove That: Cos 4 X − Cos 4 α = 8 ( Cos X − Cos α ) ( Cos X + Cos α ) ( Cos X − Sin α ) ( Cos X + Sin α )

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Question

Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]

Numerical
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Solution

\[RHS = 8\left( \text{ cos } x - cos \alpha \right) \left( \text{ cos } x + cos\alpha \right) \left( \text{ cos } x - sin\alpha \right) \left( \text{ cos } x + sin\alpha \right)\]

\[ = 8\left( \cos^2 x - \cos^2 \alpha \right) \left( \cos^2 x - \sin^2 \alpha \right)\]

\[ = 8\left( \cos^4 x - \cos^2 x \times \sin^2 \alpha - \cos^2 \alpha \times \cos^2 x + \cos^2 \alpha \times \sin^2 \alpha \right)\]

\[ = 8\left\{ \cos^4 x - \cos^2 x\left( \sin^2 \alpha + \cos^2 \alpha \right) + \cos^2 \alpha \times \sin^2 \alpha \right\}\]

\[ = 8\left\{ \cos^4 x - \cos^2 x + \cos^2 \alpha \times \left( 1 - \cos^2 \alpha \right) \right\}\]

\[ = 8\left\{ \cos^4 x - \cos^2 x + \cos^2 \alpha - \cos^4 \alpha \right\}\]

\[ = 8\left\{ \cos^2 x\left( \cos^2 x - 1 \right) + \cos^2 \alpha \times \left( 1 - \cos^2 \alpha \right) \right\}\]

\[= 8\left\{ \frac{1}{2} \cos^2 x\left( 2 \cos^2 x - 2 \right) + \frac{1}{2} \cos^2 \alpha \times \left( 2 - 2 \cos^2 \alpha \right) \right\}\]

\[ = 8\left\{ \frac{1}{2} \cos^2 x\left( 2 \cos^2 x - 1 - 1 \right) - \frac{1}{2} \cos^2 \alpha \times \left( 2 \cos^2 \alpha - 1 - 1 \right) \right\}\]

\[ = 8\left\{ \frac{1}{2} \cos^2 x\left( \cos2x - 1 \right) - \frac{1}{2} \cos^2 \alpha \times \left( \cos2\alpha - 1 \right) \right\} \left( \because \cos2\alpha = 2 \cos^2 \alpha - 1 \right) \]

\[ = 8\left[ \frac{1}{4}\left\{ 2 \cos^2 x\left( \cos2x - 1 \right) - 2 \cos^2 \alpha \times \left( \cos2\alpha - 1 \right) \right\} \right]\]

\[ = 8\left[ \frac{1}{4}\left\{ \left( 1 + \cos2x \right)\left( \cos2x - 1 \right) - \left( 1 + \cos2\alpha \right)\left( \cos2\alpha - 1 \right) \right\} \right]\]

\[= 8\left[ \frac{1}{4}\left\{ \cos^2 2x - 1 - \cos^2 2\alpha + 1 \right\} \right]\]

\[ = 8\left[ \frac{1}{8}\left\{ 2 \cos^2 2x - 2 \cos^2 2\alpha \right\} \right]\]

\[ = \left[ \left\{ \left( 1 + \cos4x \right) - \left( 1 + \cos4\alpha \right) \right\} \right] \]

\[ = \left[ 1 + \cos4x - 1 - \cos4\alpha \right]\]

\[ = \cos4x - \cos4\alpha = LHS\]

\[\text{ Hence proved }  .\]

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 28]

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R.D. Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 24 | Page 28

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