sin 47° + sin 61° − sin 11° − sin 25° is equal to
(a) sin 36°
(b) cos 36°
(c) sin 7°
(d) cos 7°

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Solution

(d) cos 7°
$\sin 47 ° + \sin 61 ° - \sin 11 ° - \sin 25 ° = \sin 47 ° - \sin 25 ° + \sin 61 ° - \sin 11 ° = 2 \sin (\frac{47 ° - 25 °}{2}) \cos (\frac{47 ° + 25 °}{2}) + 2 \sin (\frac{61 ° - 11 °}{2}) \cos (\frac{61 ° + 11 °}{2}) = 2 \sin 11 ° \cos 36 ° + 2 \sin 25 ° \cos 36 ° = 2 \cos 36 ° (\sin 11 ° + \sin 25 °) = 2 \cos 36 ° \{2 \sin (\frac{11 ° + 25 °}{2}) \cos (\frac{11 ° - 25 °}{2})\} = 4 \cos 36 ° \sin 18 ° \cos 7 ° = 4 \times (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4}) \cos 7 ° [\cos 36 ° = \frac{\sqrt{5} + 1}{4} and \sin 18 ° = \frac{\sqrt{5} - 1}{4}] = \frac{5 - 1}{4} \cos 7 ° = \cos 7 °$

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Similar questions

$s i n 47^{\circ} + s i n 61^{\circ} - s i n 11^{\circ} - s i n 25^{\circ}$ is equal to

Q. The value of

x \in (0, 90^{\circ})

satisfying

c o s x = s i n 61^{\circ} + s i n 47^{\circ} - s i n 25^{\circ} - s i n 11^{\circ}

, where

sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}

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Q. Find the value of

sin 47 ° + sin 61 ° - sin 11 ° - sin 25 °

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(iv) $c o s 80^{\circ} c o s 20^{\circ} + s i n 18^{\circ} s i n 20^{\circ}$

Q. Evaluate the following:

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sin 47° + sin 61° − sin 11° − sin 25° is equal to (a) sin 36° (b) cos 36° (c) sin 7° (d) cos 7°

sin 47° + sin 61° − sin 11° − sin 25° is equal to
(a) sin 36°
(b) cos 36°
(c) sin 7°
(d) cos 7°