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.

Both of them are positive, so let's square both of them:
first one will give: 
\begin{gather}
\frac{1}{4} \left( 4  - \sqrt{3} + 2 \sqrt{4-2\sqrt{3}}\right) = \frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{1-2\sqrt{3}+3}\right) =\frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{(1-\sqrt{3})^2}\right) = \frac{1}{4} \left( 4 - \sqrt{3} + 2 (\sqrt{3}-1) \right) =  \frac{1}{4} \left( 2 + \sqrt{3}  \right) = 0.5 + \frac{\sqrt{3}}{4}
\end{gather}
the second one:
$$\frac{1}{16} \left( 6+2+2\sqrt{12}\right) = 0.5+\frac{\sqrt{3}}{4} $$
