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Prove that: cos 4 x = 1 − 8 cos 2 x + 8 cos 4 x

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Question

Prove that:  \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]

 

Numerical
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Solution 1

\[LHS = \cos4x\]

\[ = \cos\left( 2 \times 2x \right)\]

\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]

\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]

\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]

\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]

\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]

\[Hence proved .\]

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Solution 2

\[LHS = \cos4x\]

\[ = \cos\left( 2 \times 2x \right)\]

\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]

\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]

\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]

\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]

\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]

\[\text{ Hence proved } .\]

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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 17 | Page 28

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