13^{} 17^{} 21^{} 47^{}= [AP EAPCET 24 May 2025 S1]

Correct Answer is : 132(1+23−5)\; 1/32(1 + 2{3} - {5})321(1+23−5)

Correct Answer is : 132(1+23−5)\; 1/32(1 + 2{3} - {5})321(1+23−5)

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Trigonometric Ratios and Identities

Section: Mathematics

Question : 2 of 50

Marks: +1, -0

\cos 13^{\circ} \sin 17^{\circ} \sin 21^{\circ} \cos 47^{\circ}=

Solution:

\cos 13^{\circ} \cdot \sin 17^{\circ} \cdot \sin 21^{\circ} \cdot \cos 47^{\circ}

= (\cos 13^{\circ} \cos 47^{\circ}) \cdot (\sin 17^{\circ} \sin 21^{\circ})

= \frac{1}{2} [\cos(34^{\circ}) + \cos(60^{\circ})] \cdot \frac{1}{2} [\cos(4^{\circ}) - \cos(38^{\circ})]

\; \left[ \because \sin A \sin B = \frac{1}{2}[\cos(A-B)-\cos(A+B)] \;\text{ and }\; \cos A \cos B = \frac{1}{2}[\cos(A-B)+\cos(A+B)].. \right]

= \frac{1}{4} \left[\cos 34^{\circ}+\frac{1}{2}\right] [\cos 4^{\circ}-\cos 38^{\circ}]

= \frac{1}{4} \left[\cos 34^{\circ} \cos 4^{\circ}-\cos 34^{\circ} \cos 38^{\circ}+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right]

= \frac{1}{4} \left[ \frac{1}{2} \{\cos(30^{\circ})+\cos(38^{\circ})\} - \frac{1}{2} \{\cos(4^{\circ})+\cos(72^{\circ})\} + \frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ} \right]

= \frac{1}{4} \left[ \frac{1}{2} \left(\frac{\sqrt{3}}{2}+\cos 38^{\circ}\right) - \frac{1}{2} (\cos 4^{\circ}+\cos 72^{\circ}) + \frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ} \right]

= \frac{1}{8} \left[\frac{\sqrt{3}}{2}-\cos 72^{\circ}\right] = \frac{\sqrt{3}}{16}-\frac{\cos 72^{\circ}}{8}

= \frac{\sqrt{3}}{16}-\frac{\sqrt{5}-1}{4} \times \frac{1}{8} \left( \because \cos 72^{\circ} = \frac{\sqrt{5}-1}{4}\right)

= \frac{\sqrt{3}}{16}-\frac{\sqrt{5}-1}{32} = \frac{2\sqrt{3} - \sqrt{5} + 1}{32} = \frac{1}{32}(2\sqrt{3} - \sqrt{5} + 1)

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