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

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles3 Marks
Question
prove that:
$\cos^6\text{x}=\cos2\text{x}=\cos2\text{x}(1-\frac{1}{4}\sin^22\text{x})$
✓

Answer

$\text{LHS}=\cos^6\text{x}-\sin^6\text{x}$
$=(\cos^2\text{x})^3-(\sin^2\text{x})^3$
$=(\cos^2\text{x}-\sin^2\text{x})(\cos^4\text{x}+\sin^2\text{x}.\cos^2\text{x}+\sin^4\text{x})$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\big]$
$=\cos2\text{x}(\cos^4\text{x}+2\sin^2+\sin^4\text{x}-\sin^2\text{x}\cos^2\text{x})$
$\therefore\cos^2\text{x}-\sin^2\text{x}=\cos^2\text{x}=\&$ adding and subtracling $\sin^2\text{x}\cos^2\text{x}$
$=\cos2\text{x}\big[(\sin^2\text{x}+\cos^2\text{x})^2-\frac{4}{4}\sin^2\text{x}\cos^2\text{x}$
$=\cos2\text{x}\big[1-\frac{1}{4}(2\sin\text{A}\cos\text{x})^2\big]$
$=\cos2\text{x}\big[1-\frac{1}{4}\sin^22\text{x}\big]=\text{RHS}$

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