<s>[INST] 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.
 [/INST] mathisfun did a nice job of showing you how to reconcile your answers. 
Another way of showing that $\sin(45^\circ) + \sin(15^\circ) = \sin(75^\circ)$ is to use the formula 
$$\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)$$
with $\alpha = 45^\circ$ and $\beta = 15^\circ$. 
\begin{align*}
\sin(45^\circ) + \sin(15^\circ) & = 2\sin\left(\frac{45^\circ + 15^\circ}{2}\right)\cos\left(\frac{45^\circ - 15^\circ}{2}\right)\\
& = 2\sin\left(\frac{60^\circ}{2}\right)\cos\left(\frac{30^\circ}{2}\right)\\
& = 2\sin(30^\circ)\cos(15^\circ)\\
& = 2 \cdot \frac{1}{2} \cdot \cos(15^\circ)\\
& = \cos(15^\circ)\\
& = \sin(90^\circ - 15^\circ)\\
& = \sin(75^\circ)
\end{align*}
since $\cos\theta = \sin(90^\circ - \theta)$.   
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