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Standard XII

Mathematics

Many-One onto Function

cos 6∘sin 24∘...

Question

$cos 6^{\circ} sin 24^{\circ} cos 72^{\circ} =$

- \frac{1}{8}

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- \frac{1}{4}

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\frac{1}{8}

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\frac{1}{4}

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Solution

The correct option is C

\frac{1}{8}

=cos6.sin24cos72
=cos6.sin24.cos(90-18)
=1/2[2cos6.sin24}(sin18)
=1/2[ sin(6+24)-sin(6-24)]sin18
=1/2[sin30.sin18+sin18.sin18]
=1/2[1/2*(5^1/2-1)/2) + {(5^1/2-1)/2}^2]
After simplyfing we get,
=1/8

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Similar questions

Q. Match the following.
List - I List - II
1.

cos 6^{\circ} sin 24^{\circ} cos 72^{\circ}

\frac{3}{16}

cos 10^{\circ} cos 30^{\circ} cos 50^{\circ} cos 70^{\circ}

\frac{1}{16}

sin 2^{\circ} sin 24^{\circ} sin 48^{\circ} sin 84^{\circ}

\frac{3}{16}

sin 20^{\circ} sin 40^{\circ} sin 60^{\circ} sin 80^{\circ}

\frac{1}{8}

Q. Match the following.
List - I List - II
1.

sin 48^{\circ} sin 12^{\circ}

\frac{\sqrt{3}}{\sqrt{2}}

cos 48^{\circ} cos 12^{\circ}

\frac{\sqrt{15} + \sqrt{3}}{4}

sin 75^{\circ} + sin 15^{\circ}

\frac{\sqrt{5} - 1}{8}

sin 24^{\circ} + cos 6^{\circ}

\frac{3 + \sqrt{5}}{8}

Factorise: $1 - 16 (x - 2 y)^{2}$ .

4 (sin 24^{\circ} + cos 6^{\circ}) = \sqrt{15} + \sqrt{3}

II:

sin 21^{\circ} cos 9^{\circ} - cos 84^{\circ} cos 6^{\circ} = \frac{1}{4}

Q. Evaluate:
(i)

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

(ii)

{(\frac{5}{8})}^{- 7} \times {(\frac{8}{5})}^{- 4}

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cos6∘sin24∘cos72∘=

The correct option is C 18=cos6.sin24cos72=cos6.sin24.cos(90-18)=1/2[2cos6.sin24}(sin18)=1/2[ sin(6+24)-sin(6-24)]sin18=1/2[sin30.sin18+sin18.sin18]=1/2[1/2*(5^1/2-1)/2) + {(5^1/2-1)/2}^2]After simplyfing we get,=1/8

$cos 6^{\circ} sin 24^{\circ} cos 72^{\circ} =$

The correct option is C $\frac{1}{8}$
=cos6.sin24cos72
=cos6.sin24.cos(90-18)
=1/2[2cos6.sin24}(sin18)
=1/2[ sin(6+24)-sin(6-24)]sin18
=1/2[sin30.sin18+sin18.sin18]
=1/2[1/2*(5^1/2-1)/2) + {(5^1/2-1)/2}^2]
After simplyfing we get,
=1/8