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.
 A: 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:
$$\frac{1}{16} \left( 6+2+2\sqrt{12}\right) = 0.5+\frac{\sqrt{3}}{4} $$
A: 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...
A: 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 $\cos\theta = \sin(90^\circ - \theta)$.   
A: 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?
