The correct option is D 1/2 2sin(5π/12)sin(π12)=2sin(5π/12)cos(π2 π12)=2sin(5π/12)cos(5π12) = sin(10π12) = sin5π6 = sin(π nbsp;5π6 ...

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

2sin5π/12sinπ...

Question

$2 sin (\frac{5 π}{12}) sin (\frac{π}{12}) =$

- \frac{1}{2}

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\frac{1}{2}

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\frac{1}{4}

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\frac{1}{6}

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Solution

The correct option is D $\frac{1}{2}$
$2 sin (\frac{5 π}{12}) sin (\frac{π}{12}) = 2 sin (\frac{5 π}{12}) cos (\frac{π}{2} - \frac{π}{12}) = 2 sin (\frac{5 π}{12}) cos (\frac{5 π}{12})$
$= sin (\frac{10 π}{12})$
$= sin \frac{5 π}{6}$
$= sin (π - \frac{5 π}{6})$
$= \frac{1}{2}$

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Similar questions

Prove that:
(i) $2 s i n \frac{5 π}{12} s i n \frac{π}{12} = \frac{1}{2}$
(ii) $2 c o s \frac{5 π}{12} c o s \frac{π}{12}$
(iii) $2 s i n \frac{5 π}{12} c o s \frac{5 π}{12} = \frac{\sqrt{3} + 2}{2}$

Q. Which among

2^{\frac{1}{2}}, 3^{\frac{1}{3}}, 4^{\frac{1}{4}}, 6^{\frac{1}{6}}

and

12^{\frac{1}{12}}

is the highest?

Q. If

{(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{12} x^{12}

, then

a_{2} + a_{4} + a_{6} + . . . + a_{12} =

Q. Find the value of each of the following:
(i)

{(\frac{1}{2})}^{- 1} + {(\frac{1}{3})}^{- 1} + {(\frac{1}{4})}^{- 1}

(ii)

{(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2}

(iii) (2⁻¹ × 4⁻¹) ÷ 2⁻²
(iv) (5⁻¹ × 2⁻¹) ÷ 6⁻¹

Q. Convert the following data into grouped frequency table:

12, 1, 2, 3, 4, 4, 4, 5, 6, 3, 2, 1, 12, 4, 6, 7, 7, 4

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2sin(5π12)sin(π12)=

The correct option is D 122sin(5π12)sin(π12)=2sin(5π12)cos(π2−π12)=2sin(5π12)cos(5π12)=sin(10π12)=sin5π6=sin(π−5π6)=12

$2 sin (\frac{5 π}{12}) sin (\frac{π}{12}) =$

The correct option is D $\frac{1}{2}$
$2 sin (\frac{5 π}{12}) sin (\frac{π}{12}) = 2 sin (\frac{5 π}{12}) cos (\frac{π}{2} - \frac{π}{12}) = 2 sin (\frac{5 π}{12}) cos (\frac{5 π}{12})$
$= sin (\frac{10 π}{12})$
$= sin \frac{5 π}{6}$
$= sin (π - \frac{5 π}{6})$
$= \frac{1}{2}$