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

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles1 Mark
Question
Write the angled triangle ABC, write value of $\sin^2\text{A}\sin^2\text{B}+\sin^2\text{C}.$
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Answer

$\cos^276^\circ+\cos^216^\circ-\cos76^\circ\cos16^\circ$
$=\frac{1}{2}[1+\cos2(76)^\circ+1\cos2(16)^\circ-\cos(76-16)^\circ]$
$=\frac{1}{2}\big[2-\frac{1}{2}+\cos152^\circ+\cos32^\circ-\cos92^\circ\big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin\frac{92+32}{2}\sin\frac{92-32}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin\frac{124^\circ}{2}\sin\frac{60^\circ}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin62^\circ\sin30^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin62^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin(90-28)^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+\cos28^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+2\cos\frac{152-28}{2}\cos\frac{152-28}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+2\cos90^\circ\cos\frac{124^\circ}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+2(0)\cos62^\circ\Big]$
$=\frac{3}{4}$

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