Sin 47-sin 25-sin 61-sin 11-

The correct option is A cos 7^∘ (sin 47^∘ + sin 61^∘) (sin 25^∘ + sin 11^∘) = 2 sin 54^∘cos 7^∘ sin 18^∘cos 7^∘ [Using sin C+sin D=2sin(C+D/2)cos(C D/2)] = ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

sin 47∘-sin 2...

Question

$sin 47^{\circ} - sin 25^{\circ} + sin 61^{\circ} - sin 11^{\circ} =$

$cos 7^{\circ}$

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$sin 7^{\circ}$

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$2 cos 7^{\circ}$

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$2 sin 7^{\circ}$

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Solution

The correct option is A
$cos 7^{\circ}$

$(sin 47^{\circ} + sin 61^{\circ}) - (sin 25^{\circ} + sin 11^{\circ})$

$= 2 sin 54^{\circ} cos 7^{\circ} - sin 18^{\circ} cos 7^{\circ}$ $[Using sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2})]$

$= 2 cos 7^{\circ} (sin 54^{\circ} - sin 18^{\circ})$

$= 2 cos 7^{\circ} (cos 36^{\circ} - sin 18^{\circ})$

$= 2 cos 7^{\circ} (\frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4}) = \frac{2 cos 7^{\circ}}{2} = cos 7^{\circ}$

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Similar questions

Q. Find the value of

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

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

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}

$cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$

Q. Prove that

\frac{cos 11^{\circ} + sin 11}{cos 11^{\circ} - sin 11^{\circ}} = tan 56^{\circ}

{(\frac{\sin 47 °}{\cos 43 °})}^{2} + {(\frac{\cos 43 °}{\sin 47 °})}^{2} - 4 \cos^{2} 45 °

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sin47∘−sin25∘+sin61∘−sin11∘=

The correct option is A cos7∘ (sin47∘+sin61∘)−(sin25∘+sin11∘) =2sin54∘cos7∘−sin18∘cos7∘ [Using sinC+sinD=2sin(C+D2)cos(C−D2)] =2cos7∘(sin54∘−sin18∘) =2cos7∘(cos36∘−sin18∘) =2cos7∘(√5+14−√5−14)=2cos7∘2=cos7∘

$sin 47^{\circ} - sin 25^{\circ} + sin 61^{\circ} - sin 11^{\circ} =$