Prove that: \[ \frac{ \sin A\sin2A+\sin3A\sin6A }{ \sin A\cos2A+\sin3A\cos6A } = \tan5A \]
Solution
L.H.S.
\[ = \frac{ \sin A\sin2A+\sin3A\sin6A }{ \sin A\cos2A+\sin3A\cos6A } \]Use identity:
\[ \sin C\sin D = \frac12[\cos(C-D)-\cos(C+D)] \]
\[
=
\frac{
\frac12(\cos A-\cos3A)+\frac12(\cos3A-\cos9A)
}{
\sin A\cos2A+\sin3A\cos6A
}
\]
\[
=
\frac{
\frac12(\cos A-\cos9A)
}{
\sin A\cos2A+\sin3A\cos6A
}
\]
Use identity:
\[ \cos C-\cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \]
\[
=
\frac{
\frac12(2\sin5A\sin4A)
}{
\sin A\cos2A+\sin3A\cos6A
}
\]
\[
=
\frac{
\sin5A\sin4A
}{
\sin A\cos2A+\sin3A\cos6A
}
\]
Use identity:
\[ \sin C\cos D = \frac12[\sin(C+D)+\sin(C-D)] \]
\[
=
\frac{
\sin5A\sin4A
}{
\frac12(\sin3A-\sin A)+\frac12(\sin9A-\sin3A)
}
\]
\[
=
\frac{
\sin5A\sin4A
}{
\frac12(\sin9A-\sin A)
}
\]
Use identity:
\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \]
\[
=
\frac{
\sin5A\sin4A
}{
\frac12(2\cos5A\sin4A)
}
\]
\[
=
\frac{
\sin5A\sin4A
}{
\cos5A\sin4A
}
\]
\[
=
\frac{\sin5A}{\cos5A}
\]
Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \]
\[
=
\tan5A
\]
Hence Proved.