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Trigonometric Ratios Of Compound Angles — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compound Angles5 Marks
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$3(\sin\text{x}-\cos\text{x})+6(\sin\text{x}+\cos)^2+4(\sin^6\text{x}+\cos^6\text{x})=13$
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Answer

$\text{LHS}=3(\sin\text{x}-\cos\text{x})+6(\sin\text{x}+\cos)^2+4(\sin^6\text{x}+\cos^6\text{x})$
$=3[\sin^4\text{x}-4\sin^3\text{x}\cos\text{x}+6\sin^2\text{x}\cos^2\text{x}-4\sin\text{x}\cos^3\text{x}+\cos^4\text{x}]$
$+6[\sin^2\text{x}+2\sin\text{x}\cos\text{x}+\cos^2\text{x}]+4(\sin^6\text{x}+\cos^6\text{x})$
$\big[\because(\text{a}-\text{b})^4=\text{a}^4-4\text{a}^3\text{b}+6\text{a}^2\text{b}^2-4\text{ab}^3+\text{b}^4$ by binomial expainsion$\big]$
$=3\big[\sin^4\text{x}+\cos^4\text{x}-4\sin\text{x}\cos\text{x}(\sin^2\text{x}+\cos^2\text{x})+6\sin^2\text{x}\cos^2\text{x}\big]$
$+6[1+2\sin\text{x}\cos\text{x}]+4\big[(\cos^2\text{x}+\sin^2\text{x})(\cos^4\text{x}-\cos^2\text{x}+\sin^4\text{x})\big]$
$\big[\because\text{a}^3+\text{b}^3=(\text{a+b})(\text{a}^2-\text{ab+b}^2)$
$=7[\sin^4\text{x}+\cos^4\text{x}]+18\sin^2\text{x}\cos^4\text{x}-4\sin^2\text{x}\cos^2\text{x}+6$
$=7[\sin^4\text{x}+\cos^4\text{x}2\sin^2​​\cos^2​​]+6$
$=7[\sin^2\text{x}+\cos^2\text{x}+2\sin^2\text{x}\cos^2\text{x}]+6$
$=7+6$
$=13=\text{RHS}$

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