Q:

Compound angle formulas

I need help understanding why how the textbook got a certain answer but I got a completly different answer. 
The question was to apply a compound angle formula, and then determine and exact value for each.  It's two questions (a&b) but I think that if someone can help me understand one I'll get the other. 
a) $\sin(\pi/3 + \pi/4)$
The back of the text book says $\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$
I got -0.970535282
My method was:
$$
\begin{align*}
\sin(x+y) &= \sin x  \cos y  + \cos x \sin y \\
          &= \sin \frac{\pi}{3} \cos \frac{\pi}{4} 
               + \cos \frac{\pi}{3} \sin \frac{\pi}{4} \\
          &= \sin 60º \cos 45º + \cos 60º \sin 45º \\
          &= - 0.160123721 + (- 0.810411561) \\
          &= - 0.970535282
\end{align*}
$$
EDIT:
So so far after all the help I have recieved i have done
$$
\begin{align*}
\sin(x+y) &= \sin x  \cos y  + \cos x \sin y \\
          &= \sin \frac{\pi}{3} \cos \frac{\pi}{4} 
               + \cos \frac{\pi}{3} \sin \frac{\pi}{4} \\
          &=(\dfrac{\sqrt{3}}{4} * \dfrac{\sqrt{2}}{4})  + (\dfrac1{2} * \dfrac{\sqrt{2}}        {2})\\
          &=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\
          &=\dfrac{\sqrt{3}+1}{2 \sqrt{2}}\\
\end{align*}
$$
But I'm a bit confused as to what i have to do next to get $\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$
EDIT 2:
I get it finally thanks for all the help! 
Basically have to simplify 
$\dfrac{\sqrt{6}+\sqrt{2}}{4}$
(simplify by taking out $\sqrt{2}$ )
$\sqrt{6} = \dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$
$\sqrt{2} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
$4 = \dfrac{4\sqrt{2}}{\sqrt{2}\sqrt{2}} = \dfrac{4\sqrt{2}}{2} =  2\sqrt{2}$

A:

Hint: use the formula $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ and the values given in the book for the trigonometric functions of $\pi/3$ and $\pi/4$.
Suggestion: The answer sought in most books is exact rather than plugging into a calculator.  Usually books have a table of trigonometric functions for some special values, such as π/3 and π/4.  Find those and use them. 

A:

Do you know the formula $\sin(a+b)=\sin a \cos b + \cos a \sin b?$  You should be able to apply that here. 
Your answer is close to the negative of the correct one.   If you don't apply the angle sum formula, you get $\sin \frac {7\pi}{12}$.  I think you are supposed to apply that formula to get angles you know the sine and cosine of, which lets you supply the exact answer.