The value of 8sin- 48 -cos- 48 -cos- 24 -cos- 12 is equal to-

The correct option is A 12 8sinπ #160; / 48 #160; cosπ #160; / 48 #160; cosπ #160; / 24 #160; cosπ #160; / 12 #160; =4sinπ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

The value of ...

Question

The value of $8 sin \frac{π}{48} . cos \frac{π}{48} . cos \frac{π}{24} . cos \frac{π}{12}$ is equal to-

\frac{1}{2}

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

- \frac{1}{2}

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

\frac{1}{4}

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

\frac{1}{8}

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

Open in App

Solution

Suggest Corrections

Similar questions

(c o s \frac{π}{8} + i s i n \frac{π}{8}) \times (c o s \frac{π}{12} + i s i n \frac{π}{12}) \times (c o s \frac{π}{24} + i s i n \frac{π}{24}) \times (c o s \frac{π}{4} + i s i n \frac{π}{4})

{(\frac{1 + c o s \frac{π}{2} - i s i n \frac{π}{2}}{1 + c o s \frac{π}{2} + i s i n \frac{π}{2}})}^{2}

{⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 + cos (\frac{π}{12}) + i sin (\frac{π}{12})}{1 + cos (\frac{π}{12}) - i sin (\frac{π}{12})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{72} =

{(\frac{1 + cos \frac{π}{8} - i sin \frac{π}{8}}{1 + cos \frac{π}{8} - i sin \frac{π}{8}})}^{2} =

{⎛ ⎜ ⎜ ⎝ \frac{1 + cos \frac{π}{8} - i sin \frac{π}{8}}{1 + cos \frac{π}{8} + i sin \frac{π}{8}} ⎞ ⎟ ⎟ ⎠}^{8} =

Join BYJU'S Learning Program