prove that $\sqrt{2} \sin10^\circ+ \sqrt{3} \cos35^\circ= \sin55^\circ+ 2\cos65^\circ$ Question:
Prove that: $\sqrt{2} \sin10^\circ + \sqrt{3} \cos35^\circ = \sin55^\circ + 2\cos65^\circ$
My Efforts:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$

As Deepak wrote it's easier to start with the right hand side, but you could go ahead with your approach too, as shown by Arian.  
\begin{equation*}
\sqrt{2}\sin 10{{}^\circ}
+\sqrt{3}\cos 35{{}^\circ}
=\sin 55{{}^\circ}+2\cos 65{{}^\circ}\tag{1}
\end{equation*}
Here is a variant that uses the cosine addition formula for $65{{}^\circ}=35{{}^\circ}+30{{}^\circ}$, the sine difference formula for $10{{}^\circ}=45{{}^\circ}-35{{}^\circ}$ and the sine/cosine complement formula.  


*

*Use the complement formula $\sin \theta =\cos \left( 90{{}^\circ}
-\theta \right) $ for $\theta=55{{}^\circ}$, the addition formula $\cos \left( a+b\right) =\cos a\cos
b-\sin a\sin b,$ and the special values $\sin 30{{}^\circ}
=\frac{1}{2}$, $\cos 30{{}^\circ}
=\frac{\sqrt{3}}{2}$ to rewrite the right hand side of $(1)$ as [hover your mouse over the grey to see]

\begin{equation*}\sin 55{{}^\circ}+2\cos 65{{}^\circ}=\cos 35{{}^\circ}+\sqrt{3}\cos 35{{}^\circ}-\sin 35{{}^\circ}\tag{2}\end{equation*}



*Substitute $(2)$ in $(1)$ and simplify to obtain the equivalent identity 



\begin{equation*}\sqrt{2}\sin 10{{}^\circ}=\cos 35{{}^\circ}-\sin 35{{}^\circ}\tag{3}\end{equation*}



*To show that $(3)$ holds, use the subtraction formula $\sin \left(
a-b\right) =\sin a\cos b-\cos a\sin b$ and the special values $\sin 45{{}^\circ}=\cos 45{{}^\circ}=\frac{\sqrt{2}}{2}$. Since we get the following trivial identity, we are
done. 



\begin{equation*}\cos 35{{}^\circ}-\sin 35{{}^\circ}=\cos 35{{}^\circ}-\sin 35{{}^\circ}\tag{4}\end{equation*}

