$sin A + sin 3 A + sin 5 A + sin 7 A$ equals

4 sin A cos 2 A cos 4 A

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

4 sin A cos 2 A cos 2 A

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

4 cos A sin 2 A sin 4 A

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

4 cos A cos 2 A sin 4 A

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

Open in App

Solution

The correct option is D $4 cos A cos 2 A sin 4 A$
$sin A + sin 3 A + sin 5 A + sin 7 A$
$= 2 sin 2 A cos A + 2 sin 6 A cos A$
$= 2 [sin 2 A + sin 6 A] cos A$
$= 2 [2 sin 4 A cos 2 A] cos A$
$= 4 sin 4 A cos A cos 2 A$
$= 4 cos A cos 2 A sin 4 A$

Suggest Corrections

Similar questions

Prove that sin A - sin 3A + sin 5A - sin 7A/ sin A - sin 3A - sin 5A +sin 7A = cot 2A.

\frac{sin 3 A + 2 sin 5 A + sin 7 A}{sin A + 2 sin 3 A + sin 5 A} = \frac{sin 5 A}{sin 3 A}

sin A + sin 3 A + sin 5 A + sin 7 A =

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A}

\frac{c o s A + c o s 3 A + c o s 5 A + c o s 7 A}{s i n A + s i n 3 A + s i n 5 A + s i n 7 A} =

Join BYJU'S Learning Program