- r -116-1-cos 4 r - where -r- is in degreesA- 4 -2sin 1-cos 2-cos 17- 4 -2cos 1-cos 2-cos 17-C- 4 -2cos 1-sin 2-cos 17-D- -2sin 16-cos 17-sin 1-

The correct option is D √(2)sin 16^∘cos 17^∘/sin 1^∘∑_r=1^16√(1+cos 4r) =∑_r=1^16√(2) cos 2r =√(2)(cos 2^∘+cos 4^∘+cos 6^∘+....+cos 32^∘) =√(2)sin 16^∘/sin 1 ...

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Byju's Answer

Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

∑ r =116√1+co...

Question

$\sum_{r = 1}^{16} \sqrt{1 + c o s 4 r}$ =(where 'r' is in degrees)

4 \sqrt{2} s i n 1^{\circ} c o s 2^{\circ} c o s 17^{\circ}

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4 \sqrt{2} c o s 1^{\circ} c o s 2^{\circ} c o s 17^{\circ}

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4 \sqrt{2} c o s 1^{\circ} s i n 2^{\circ} c o s 17^{\circ}

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\sqrt{2} \frac{s i n 16^{\circ} c o s 17^{\circ}}{s i n 1^{\circ}}

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Solution

The correct option is D $\sqrt{2} \frac{s i n 16^{\circ} c o s 17^{\circ}}{s i n 1^{\circ}}$
$\sum_{r = 1}^{16} \sqrt{1 + c o s 4 r} = \sum_{r = 1}^{16} \sqrt{2} c o s 2 r = \sqrt{2} (c o s 2^{\circ} + c o s 4^{\circ} + c o s 6^{\circ} + . . . . + c o s 32^{\circ}) = \sqrt{2} \frac{s i n 16^{\circ}}{s i n 1^{\circ}} c o s {2 + (15) (1)} = \sqrt{2} \frac{s i n 16^{\circ} c o s 17^{\circ}}{s i n 1^{\circ}}$

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Similar questions

$\sum_{r = 1}^{16} \sqrt{1 + c o s 4 r} =$

\frac{cos 17^{\circ} + sin 17^{\circ}}{cos 17^{\circ} - sin 17^{\circ}}

Q. The value of the expression

\frac{2 (sin 1^{\circ} + sin 2^{\circ} + sin 3^{\circ} + . . . + sin 89^{\circ})}{2 (cos 1^{\circ} + cos 2^{\circ} + . . . + cos 44^{\circ}) + 1}

(\frac{sin 49^{\circ}}{cos 41^{\circ}} - \frac{cos 17^{\circ}}{sin 73^{\circ}}) =

P r o v e t h a t : \frac{cos 17^{\circ} + sin 17^{\circ}}{cos 17^{\circ} - sin 17^{\circ}} = tan 62^{\circ}

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∑16r=1√1+cos4r =(where 'r' is in degrees)

The correct option is D √2sin16∘cos17∘sin1∘∑16r=1√1+cos4r=∑16r=1√2cos2r=√2(cos2∘+cos4∘+cos6∘+....+cos32∘)=√2sin16∘sin1∘cos{2+(15)(1)}=√2sin16∘cos17∘sin1∘

$\sum_{r = 1}^{16} \sqrt{1 + c o s 4 r}$ =(where 'r' is in degrees)