Question

sin7 Θ +sin 3Θ convert into a form of product

Answer 1

Hey friend here is your answer
we know a formula which is sinc+sinD=2sinC+D/2.cosC-D/2
comparing from formula c=7a &D=3a
so using formula,
=2sin7a+3a/2.cos7a-3a/2
=2sin5a.cos2a
Hope it will help you

Answer 2

Answer is: $2\sin 5\theta \cos 2\theta$

Here we will use the identity:

$\boxed{\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)}$

Here, We can take:
$C =7\theta \\ \\ D = 3\theta$

And so, the question can be solved as follows:

$\sin 7\theta + \sin 3\theta \\ \\ \\ = 2 \sin \left(\frac{7\theta + 3\theta}{2} \right) \cos \left( \frac{7\theta - 3\theta}{2} \right) \\ \\ \\ = 2 \sin \left( \frac{10\theta}{2} \right) \cos \left( \frac{4\theta}{2}\right) \\ \\ \\ = 2 \sin 5\theta \cos 2\theta \\ \\ \\ \\ \implies \boxed{\sin 7\theta + \sin 3\theta = 2 \sin 5\theta \cos 2\theta}$