$sin 12^{o} sin 48^{o} sin 54^{o}$ is equal to ?

\frac{1}{8}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{7}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{8}$
$sin 12^{o} sin 48^{o} sin 54^{o} = \frac{1}{2} (2 sin 12^{o} sin 48^{o}) sin 54^{o} = \frac{1}{2} (cos 36^{0} - cos 60^{0}) sin 54^{o} (2 sin A sin B = cos (A - B) - cos (A + B)) = \frac{1}{2} (\frac{\sqrt{5} + 1}{4} - \frac{1}{2}) (\frac{\sqrt{5} + 1}{4}) [∵ sin 54^{o} = sin (90^{o} - 36^{0}) = cos 36^{o} = \frac{\sqrt{5} + 1}{4}] = \frac{1}{2} (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4}) = \frac{1}{8}$

Suggest Corrections

Similar questions

$\frac{1}{(\sqrt{9} - \sqrt{8})} - \frac{1}{(\sqrt{8} - \sqrt{7})} + \frac{1}{(\sqrt{7} - \sqrt{6})} - \frac{1}{(\sqrt{6} - \sqrt{5})} + \frac{1}{(\sqrt{5} - \sqrt{4})}$ is equal to:

Q. The value of

\frac{1}{\sqrt{9} - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - \sqrt{4}}

is equal to

$Name the multiplication property of rational numbers shown below: (i) \frac{3}{5} \times \frac{- 8}{9} = \frac{- 8}{9} \times \frac{3}{5} (i i) \frac{- 3}{4} \times (\frac{5}{7} \times \frac{- 8}{15}) = (\frac{- 3}{5} \times \frac{5}{7}) \times \frac{- 8}{15} (i i i) \frac{4}{5} \times (\frac{3}{- 8} + \frac{- 4}{7}) = \frac{4}{5} \times \frac{3}{- 8} + \frac{4}{5} \times \frac{- 4}{7} (i v) \frac{- 7}{5} \times \frac{5}{- 7} = 1 (v) \frac{8}{- 9} \times 1 = 1 \times \frac{8}{- 9} = \frac{8}{- 9} (v i) \frac{- 3}{4} \times 0 = 0$

Q. In the given series
1, 8, 3, 5, 1, 3, 7, 8, 1, 5, 7, 3, 5, 8, 3, 1, 5, 3, 7, 1, 8, 3, 5, 3, 7, 8, 3, 5, 1, 7, 3, 7, 5, 1, 8, 3, 1, 7, 1, 8, 3, 8, 5, 1

How many times do the two consecutive (i.e. numbers one after the other) have a difference of

4

In the given series

1, 8, 3, 5, 1, 3, 7, 8, 1, 5, 7, 3, 5, 8, 3, 1, 5, 3, 7, 1, 8, 3, 5, 3, 7, 8, 3, 5, 1, 7, 3, 7,

5, 1, 8, 3, 1, 7, 1, 8, 3, 8, 5, 1

How many times the two consecutive numbers (i.e. numbers one after the other) are such that the first is

4

more than the second?

Join BYJU'S Learning Program

sin12osin48osin54o is equal to ?

The correct option is A 18sin12osin48osin54o=12(2sin12osin48o)sin54o=12(cos360−cos600)sin54o(2sinAsinB=cos(A−B)−cos(A+B))=12(√5+14−12)(√5+14)[∵sin54o=sin(90o−360)=cos36o=√5+14]=12(√5−14)(√5+14)=18

$sin 12^{o} sin 48^{o} sin 54^{o}$ is equal to ?