Prove that: \[ \sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin2x\sin5x \]
Solution
Using the identity:
\[
\sin A \sin B
=
\frac{1}{2}
\left[
\cos(A-B)-\cos(A+B)
\right]
\]
First term:
\[
\sin\frac{x}{2}\sin\frac{7x}{2}
=
\frac{1}{2}
\left[
\cos\left(\frac{x}{2}-\frac{7x}{2}\right)
–
\cos\left(\frac{x}{2}+\frac{7x}{2}\right)
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos(-3x)-\cos4x
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos3x-\cos4x
\right]
\]
Second term:
\[
\sin\frac{3x}{2}\sin\frac{11x}{2}
=
\frac{1}{2}
\left[
\cos\left(\frac{3x}{2}-\frac{11x}{2}\right)
–
\cos\left(\frac{3x}{2}+\frac{11x}{2}\right)
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos(-4x)-\cos7x
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos4x-\cos7x
\right]
\]
Adding both:
\[
=
\frac{1}{2}
(\cos3x-\cos4x)
+
\frac{1}{2}
(\cos4x-\cos7x)
\]
\[
=
\frac{1}{2}
(\cos3x-\cos7x)
\]
Using the identity:
\[
\cos A-\cos B
=
-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
=
\frac{1}{2}
\left[
-2\sin\frac{3x+7x}{2}
\sin\frac{3x-7x}{2}
\right]
\]
\[
=
-\sin5x\sin(-2x)
\]
Since,
\[
\sin(-\theta)=-\sin\theta
\]
\[
=
\sin5x\sin2x
\]
\[
=
\sin2x\sin5x
\]
Hence,
\[
\boxed{
\sin\frac{x}{2}\sin\frac{7x}{2}
+
\sin\frac{3x}{2}\sin\frac{11x}{2}
=
\sin2x\sin5x
}
\]