Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 }  } =\sqrt { \frac { 1 }{ 2 }  } =\frac { \sqrt { 2 }  }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 }  } =\sqrt { \frac { 1-\frac { \sqrt { 3 }  }{ 2 }  }{ 2 }  } =\frac { \sqrt { 2-\sqrt { 3 }  }  }{ 2 } $$
So, $\sin45+\sin15=\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 }  }  }{ 2 } $
2) Rewriting $\sin75$ 
$$\sin75=\sin(45+30)=\sin45\cos30+\sin30\cos45$$
$$=\frac { \sqrt { 2 }  }{ 2 } \cdot \frac { \sqrt { 3 }  }{ 2 } +\frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 }  }{ 2 } $$
$$=\frac { \sqrt { 6 } +\sqrt { 2 }  }{ 4 } $$
So now I have these two expressions that are obviously equal (when I compute them) but how do I show them as being equal to each other. I assume the question is implying for the resulting expressions to look equal.

This gets easier if you compute $\sin 15^\circ$ in the same way that you computed $\sin 75^\circ$:
$$
\sin 15^\circ = \sin(45^\circ-30^\circ)=\sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ = \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \, .
$$
For an alternate approach, you could use a sum-to-product formula to note that
$$\sin 15^\circ+\sin 45^\circ = 2 \sin 30^\circ \cos 15^\circ $$
and proceed from there...
