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.

We have:
$$\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 }  }  }{ 2 }$$
By squaring the numerator and denominator:
$$\frac {\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)^2}{4} = \frac{2+2-\sqrt{3}+ 2\sqrt{4-2\sqrt3}}{4} = \frac{4-\sqrt{3}+2\sqrt{4-2\sqrt3}}{4}$$
Next we must find:
$$\sqrt{4-2\sqrt3}$$
Let's call the answer $a-\sqrt{b}$
Now, we can say that:
$$\left(\sqrt{4-2\sqrt3}\right)^2 = 4 - 2\sqrt{3} = (a-\sqrt{b})^2 = a^2 - 2a\sqrt{b} + b$$
Therefore we can say:
$$a^2 + b = 4$$
and 
$$2a\sqrt{b} = 2\sqrt{3}$$
We then get that: $a = 1$, $b = 3$
Therefore, $$\sqrt{4-2\sqrt3} = 1 - \sqrt{3}$$
Substituting:
$$\frac{4-\sqrt{3}+2\sqrt{4-2\sqrt3}}{4} = \frac{4-\sqrt{3}+2(1-\sqrt{3})}{4} = \frac{4-\sqrt{3}+2 - 2\sqrt{3}}{4} = \frac{6- 3\sqrt{3}}{4} = \frac{3(2- \sqrt{3})}{4}$$
Can you take the rest from here?
