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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:
$\sin^242^\circ-\cos^278^\circ=\frac{\sqrt{15}+1}{8}$
✓

Answer

$\text{LHS}=\sin^242^\circ-\cos^278^\circ$
$=\sin^2(90-48)-\cos^2(90-12)$
$=\cos^248^\circ-\sin^212^\circ$
$=\cos(48+12).\cos(48-12)$$\big[\because\cos(\text{A+B}).\cos(\text{A}-\text{B}=\cos^2\text{A}-\sin^2\text{B})\big]$
$=\cos60^\circ.\cos36^\circ$
$=\frac{1}{2}.\frac{\sqrt{5}+1}{4}$ $\Big[\because\cos36^\circ=\frac{\sqrt{5}+1}{4}\Big]$
$=\frac{\sqrt{5}+1}{8}$
$=\text{RHS}$

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