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:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$
 A: Hints:
$$\begin{align}&\bullet\;\;\sin(x+y)=\sin x\cos y+\sin y\cos x\\{}\\&\bullet\;\;\sin 45^\circ=\frac1{\sqrt2}=\frac{\sqrt2}2=\cos45^\circ\\{}\\&\bullet\;\;\cos x=\cos(-x)\end{align}$$
A: 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*}

A: Going from RHS to LHS is easy once you keep in mind that:
1) $\displaystyle 65^\circ = 35^\circ + 30^\circ$
2) $\displaystyle 55^\circ = 45^\circ + 10^\circ$
and finally
3) $\displaystyle 35^\circ = 45^\circ - 10^\circ$
Apply the first two using sum/difference of angles identities in the first step, and then apply the third to the result. You don't have to know anything other than the trigonometric ratios of the "special" angles $30^\circ$ and $45^\circ$.
A: $$\sqrt{2}\sin10+\sqrt{3}\cos35=2(\frac{\sqrt{2}}{2}\sin10+\frac{\sqrt{3}}{2}\cos35)=2(\sin45\sin10+\cos30\cos35)$$
using the product rule for sine and cosine yields
$$2(\sin45\sin10+\cos30\cos35)=2(\frac{\cos(45-10)-\cos(45+10)}{2}+\frac{\cos(30+35)+\cos(35-30)}{2})$$$$=\cos35-\cos55+\cos65+\cos5$$
$$=\cos(90-55)+\cos65+(\cos5-\cos55)$$
$$=\sin55+\cos65+2\sin(\frac{55+5}{2})\sin(\frac{55-5}{2})$$
$$=\sin55+\cos65+2\sin30\sin25$$
$$=\sin55+\cos65+2\frac{1}{2}\sin(90-65)$$
$$=\sin55+\cos65+\cos65$$
$$=\sin55+2\cos65$$
