Let S-1-sin 8- - 1-sin 16-1-sin 32- - - -1-sin 4096- - 1-sin 8192- If S-1-sin - where -0- 90- then - in degree is

S=∑^10_r=01/sin (2^r+3) =∑^10_r=0sin (2^r+3 2^r+2)/sin(2^r+2) sin(2^r+3) S=∑^10_r=0 [ (2^r+2) (2^r+3) ] S= ( (2^2) (2^3) )+ ( (2^3) (2^4) )+ …… + ( ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Let S=1/sin 8...

Question

Let $S = \frac{1}{sin 8^{\circ}} + \frac{1}{sin 16^{\circ}} + \frac{1}{sin 32^{\circ}} + \dots \dots + \frac{1}{sin 4096^{\circ}} + \frac{1}{sin 8192^{\circ}}$ . If $S = \frac{1}{sin α}$ , where $α \in (0, 90^{\circ})$ , then $α$ (in degree) is

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Solution

$S = 10 \sum r = 0 \frac{1}{sin (2^{r + 3})}$

$= 10 \sum r = 0 \frac{sin (2^{r + 3} - 2^{r + 2})}{sin (2^{r + 2}) sin (2^{r + 3})}$

$S = 10 \sum r = 0 [cot (2^{r + 2}) - cot (2^{r + 3})]$

$S = (cot (2^{2}) - cot (2^{3})) + (cot (2^{3}) - cot (2^{4})) + \dots \dots + (cot (2^{12}) - cot (2^{13}))$

$S = cot 4 ° - cot (8192 °)$

$= cot 4 ° + tan 2 ° = cosec (4 °)$
Here, Every angle is taken in degree.

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Similar questions

Q. Let

S = \frac{1}{sin 8^{\circ}} + \frac{1}{sin 16^{\circ}} + \frac{1}{sin 32^{\circ}} + \dots \dots + \frac{1}{sin 4096^{\circ}} + \frac{1}{sin 8192^{\circ}}

. If

S = \frac{1}{sin α}

, where

α \in (0, 90^{\circ})

, then

α

(in degree) is

$I f \frac{2 s i n α}{1 + c o s α + s i n α} = y, t h e n \frac{1 - c o s α + s i n α}{1 + s i n α} i s e q u a l t o$

Q. If

sin α, {sin}^{2} α, 1, {sin}^{4} α

and

{sin}^{5} α

are in A.P. where

- π < a < π

, then

α

lies in the interval-

Q. If

\sqrt{\frac{1 - s i n α}{1 + s i n α}} = s e c α - t a n α

, then the quadants in which

α

lies are:

Q. If

\frac{cos α}{cos θ} + \frac{sin α}{sin θ} = \frac{cos β}{cos θ} + \frac{sin β}{sin θ} = 1

where

α

and

β

do not differ by an even multiple of

π

, then show

\frac{cos α cos β}{{cos}^{2} θ} + \frac{sin α sin β}{{sin}^{2} θ} + 1 = 0

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Let S=1sin 8∘+1sin 16∘+1sin 32∘+……+1sin 4096∘+1sin 8192∘. If S=1sin α, where α∈(0,90∘), then α (in degree) is

S=10∑r=01sin(2r+3) =10∑r=0sin(2r+3−2r+2)sin(2r+2)sin(2r+3) S=10∑r=0[cot(2r+2)−cot(2r+3)] S=(cot(22)−cot(23))+(cot(23)−cot(24))+……+(cot(212)−cot(213)) S=cot 4°−cot(8192°) =cot 4°+tan 2°=cosec(4°) Here, Every angle is taken in degree.

Let $S = \frac{1}{sin 8^{\circ}} + \frac{1}{sin 16^{\circ}} + \frac{1}{sin 32^{\circ}} + \dots \dots + \frac{1}{sin 4096^{\circ}} + \frac{1}{sin 8192^{\circ}}$ . If $S = \frac{1}{sin α}$ , where $α \in (0, 90^{\circ})$ , then $α$ (in degree) is