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Show that: 3(sin⁡x−cos⁡x)4+6(sin⁡x+cos)2+4(sin6⁡x+cos6⁡x)=13 - Mathematics

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Question

Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]

Numerical
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Solution

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

`3{(sinx-cosx)^2}^2+6(sin^2x+cos^2x+2sinx.cosx)+4{(sin^2x)^3+(cos^2)^3}`

= `3(sin^2x+cos^2x-2sinxcosx)^2+6(1+2sinxcosx)+4{(sin^2x+cos^2x)((sin^2x)^2+(cos^2x)^2-sin^2x+cos^2x)}`       ...`["a"^3+"b"^3=("a"+"b")("a"^2+"b"^2-"ab")]`

= `3(1-2sinxcosx)^2+6(1+2sinxcosx)+4(sin^4x+cos^4-sin^2xcos^2x)`

= `3(1+4sin^2xcos^2x-4sinxcosx)+6+12sinxcosx+4{(sin^2x+cos^2x)^2-2cos^2xsin^2x-sin^2xcos^2x}`

= `3+12sin^2cos^2x-12sinxcosx+6+12sinxcosx+4{(sin^2x+cos^2)^2-sin^2xcos^2x}`

= `9+12sin^2cos^2x+4-12sin^2cos^2x`

= 13 = R.H.S

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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RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 19 | Page 28

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