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Show That: 2 ( Sin 6 X + Cos 6 X ) − 3 ( Sin 4 X + Cos 4 X ) + 1 = 0

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Question

Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]

 
Numerical
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Solution

\[LHS = 2\left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1\]

\[ = 2\left\{ \left( \sin^2 x \right)^3 + \left( \cos^2 \right)^3 \right\} - 3\left( \sin^4 x + \cos^4 x \right) + 1\]

\[= 2\left[ \left( \sin^2 x + \cos^2 x \right)\left( \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x \right) \right] - 3\left( \sin^4 x + \cos^4 x \right) + 1\]

\[ = 2\left( \sin^4 x + \cos^4 x \right) - 2 \sin^2 x \cos^2 x - 3\left( \sin^4 x + \cos^4 x \right) + 1\]

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

\[ = - 1 + 1 = 0 = RHS\]

\[\text{ Hence proved } .\]

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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 20 | Page 28

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