$(\sin\theta+\cos\theta)^2=1+\sin2\theta$   49)  $(\sin\theta+\cos\theta)^2=1+\sin2\theta$
Left Side:
\begin{align*}
 (\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta  
\end{align*}
This can either be $1$ or I can power reduce it. I don't know.  
Right Side:
\begin{align*}
 1+\sin2\theta=1+2\sin\theta\cos\theta  
\end{align*}
Thank you!

Remember the binomial formula. Importantly, $(a+b)^2\neq a^2+b^2$ ! Rather, 
$\begin{align*}
 (\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta = 1 + 2\sin\theta\cos\theta  
\end{align*}$
Looking at your previous questions, I think you should be a little more careful with the 'simpler' steps in your calculations and double-check those, otherwise you get lost further down.
