91Ó°ÊÓ

We are given the equation $\sin \mathrm{θ}\left[\sin \left(3\mathrm{θ}\right)+\sin \left(5\mathrm{θ}\right)\right]=\cos \mathrm{θ}\left[\cos \left(3\mathrm{θ}\right)-\cos \left(5\mathrm{θ}\right)\right]$and we need to establish it as an identity.

We first simplify the L.H.S. using the sum to product formula.

The standard formula required is

$\sin \mathrm{Î\pm }+\sin \mathrm{β}=2\sin \left(\frac{\mathrm{Î\pm }+\mathrm{β}}{2}\right)\cos \left(\frac{\mathrm{Î\pm }-\mathrm{β}}{2}\right)$.

Comparing it with the given expression in the numerator we get, $\mathrm{Î\pm }=3\mathrm{θ},\mathrm{β}=5\mathrm{θ}$.

Using the above formula we get,

$$\begin{gathered} \sin \mathrm{θ}\left[\sin \left(3\mathrm{θ}\right)+\sin \left(5\mathrm{θ}\right)\right] \\ =\sin \mathrm{θ}\left[2\sin \left(\frac{3\mathrm{θ}+5\mathrm{θ}}{2}\right)\cos \left(\frac{3\mathrm{θ}-5\mathrm{θ}}{2}\right)\right] \\ =\sin \mathrm{θ}\left[2\sin \left(\frac{8\mathrm{θ}}{2}\right)\cos \left(-\frac{2\mathrm{θ}}{2}\right)\right] \\ =\sin \mathrm{θ}\left[2\sin \left(4\mathrm{θ}\right)\cos \left(-\mathrm{θ}\right)\right] \end{gathered}$$

We know, $\cos \left(-\mathrm{θ}\right)=\cos \mathrm{θ}$.

$$\begin{gathered} ∴\sin \mathrm{θ}\left[2\sin \left(4\mathrm{θ}\right)\cos \left(-\mathrm{θ}\right)\right] \\ =\sin \mathrm{θ}\left[2\sin \left(4\mathrm{θ}\right)\cos \mathrm{θ}\right] \end{gathered}$$

It can also be written as

$\cos \mathrm{θ}\left[2\sin \left(4\mathrm{θ}\right)\sin \mathrm{θ}\right]$

The product to sum formula required here is

$\sin \mathrm{Î\pm }\sin \mathrm{β}=\frac{1}{2}\left[\cos \left(\mathrm{Î\pm }-\mathrm{β}\right)-\cos \left(\mathrm{Î\pm }+\mathrm{β}\right)\right]$

$$\begin{gathered} ∴\cos \mathrm{θ}\left[2\sin \left(4\mathrm{θ}\right)\sin \mathrm{θ}\right] \\ =\cos \mathrm{θ}\left[\frac{2}{2}\cos \left(4\mathrm{θ}-\mathrm{θ}\right)-\cos \left(4\mathrm{θ}+\mathrm{θ}\right)\right] \\ =\cos \mathrm{θ}\left[\cos \left(3\mathrm{θ}\right)-\cos \left(5\mathrm{θ}\right)\right] \end{gathered}$$

$∴\sin \mathrm{θ}\left[\sin \left(3\mathrm{θ}\right)+\sin \left(5\mathrm{θ}\right)\right]=\cos \mathrm{θ}\left[\cos \left(3\mathrm{θ}\right)-\cos \left(5\mathrm{θ}\right)\right]$

Thus, we can see that the L.H.S. of the identity is equal to the R.H.S. of the identity.

Therefore, the given equation is established as an identity.