Question

prove that sin theta/2× sin 7 theta/2 + sin 3 theta/2× sin 11 theta/2 = sin2 theta× sin5 theta

Answer 1

Here's the answer of your problem.

Answer 2

Answer:

We have given that,

$\sin{\dfrac{\theta}{2}}\sin{\dfrac{7\theta}{2}}+\sin{\dfrac{3\theta}{2}}\sin{\dfrac{11\theta}{2}}=\sin{2\theta}\sin{5\theta}$

We can take the left hand side:

$L.H.S=\sin{\dfrac{\theta}{2}}\sin{\dfrac{7\theta}{2}}+\sin{\dfrac{3\theta}{2}}\sin{\dfrac{11\theta}{2}}\\ =\dfrac{1}{2}\left[2\sin{\dfrac{\theta}{2}}\sin{\dfrac{7\theta}{2}}+2\sin{\dfrac{3\theta}{2}}\sin{\dfrac{11\theta}{2}}\right]\\=\dfrac{1}{2}\left[\cos\left(\dfrac{\theta}{2}-\dfrac{7\theta}{2}\right)-\cos\left(\dfrac{\theta}{2}+\dfrac{7\theta}{2}\right)+\cos\left(\dfrac{3\theta}{2}-\dfrac{11\theta}{2}\right)-\cos\left(\dfrac{3\theta}{2}+\dfrac{11\theta}{2}\right)\right]\\=\dfrac{1}{2}\left[\cos(-3\theta)-\cos(4\theta)+\cos(-4\theta)-\cos(7\theta)\right]\\=\dfrac{1}{2}\left[\cos(3\theta)-\cos(4\theta)+\cos(4\theta)-\cos(7\theta)\right]\\=\dfrac{1}{2}\left[\cos(3\theta)-\cos(7\theta)\right]\\=\dfrac{1}{2}\left[-2\sin\left\{\dfrac{3\theta+7\theta}{2} \right\}\sin\left\{\dfrac{3\theta-7\theta}{2}\right\}\right]\\=\dfrac{1}{2}\left[-2\sin(5\theta)\sin(-2\theta)\right]\\=\sin(5\theta)\sin(2\theta)\\=L.H.S=R.H.S$