Since \(\displaystyle \sin \alpha = \frac {1}{5}\) and \(\displaystyle \alpha\) is in quadrant I, we can say that \(\displaystyle y=1\) and \(\displaystyle r=5\) and therefore:

\(\displaystyle x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}\). 

So \(\displaystyle \cos \alpha = \frac {2\sqrt{6}}{5}\).

Since \(\displaystyle \cos \beta = \frac {1}{4}\) and \(\displaystyle \beta\) is in quadrant I, we can say that \(\displaystyle x=1\) and \(\displaystyle r=4\) and therefore: 

\(\displaystyle y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}\).  So \(\displaystyle \sin \beta = \frac {\sqrt{15}}{4}\).

Using the cosine sum formula, we then see:

\(\displaystyle \\ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\\ \\* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} - \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} - \sqrt {15}}{20}\).

Since \(\displaystyle \sin \alpha = \frac {1}{5}\) and \(\displaystyle \alpha\) is in quadrant I, we can say that \(\displaystyle y = 1\) and \(\displaystyle r = 5\) and therefore: 

\(\displaystyle x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}\). 

So \(\displaystyle \cos \alpha = \frac {2\sqrt{6}}{5}\).

Since \(\displaystyle \cos \beta = \frac {1}{4}\) and \(\displaystyle \beta\) is in quadrant I, we can say that \(\displaystyle x = 1\) and \(\displaystyle r = 4\) and therefore: 

\(\displaystyle y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}\). 

So \(\displaystyle \sin \beta = \frac {\sqrt{15}}{4}\).

Using the cosine difference formula, we see:
\(\displaystyle \\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\\ \\* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} + \sqrt {15}}{20}\)

Using the sum formula for sine,

\(\displaystyle \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\)

where, 

\(\displaystyle a=\frac{\pi}{3}\), \(\displaystyle b=\frac{\pi}{6}\)

yeilds:

\(\displaystyle \sin \bigg(\frac{\pi}{3}+\sin\frac{\pi}{6}\bigg)=\sin\bigg(\frac{\pi }{3}\bigg)\cdot\cos\bigg(\frac{\pi }{6}\bigg) +\cos \bigg(\frac{\pi}{3}\bigg)\cdot\sin\bigg(\frac{\pi }{6}\bigg)\)

\(\displaystyle =\frac{\sqrt{3}}2{}\cdot\frac{\sqrt{3}}2{}+\frac{1}{2}\cdot\frac{1}{2}\)

\(\displaystyle =\frac{3}{4}+\frac{1}{4}=1\).

Notice that \(\displaystyle \sin(75)\) is equivalent to \(\displaystyle \sin(30+45)\). With this conversion, the sum formula can be applied using,

\(\displaystyle \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\)

where

\(\displaystyle a=30\), \(\displaystyle b=45\).

Therefore the result is as follows:

 \(\displaystyle \sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)\)

\(\displaystyle =\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}2{}\)

\(\displaystyle =\frac{\sqrt2}{4}+\frac{\sqrt6}{4}=\frac{\sqrt2 +\sqrt6}{4}\). 

In order to solve this question, it is necessary to know the sine difference identity.

\(\displaystyle sin(A-B)=sinAcosB-cosAsinB\)

The values of \(\displaystyle A\) and\(\displaystyle B\) must be a special angle, and their difference must be 15 degrees.

A possibility of their values that match the criteria are:

\(\displaystyle A=45\)

\(\displaystyle B=30\)

Substitute the values into the formula and solve.

\(\displaystyle sin(45-30)=(sin45)(cos30)-(cos45)(sin30)\)

\(\displaystyle sin(15)=(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})-(\frac{\sqrt2}{2})(\frac{1}{2})= \frac{\sqrt6}{4}-\frac{\sqrt2}{4}\)

Evaluate \(\displaystyle 4sin(15)\).

\(\displaystyle =4(\frac{\sqrt6}{4}-\frac{\sqrt2}{4})= \sqrt6-\sqrt2\)

In order to find the exact value of \(\displaystyle sin(105)\), the sum identity of sine must be used.  Write the formula.

\(\displaystyle sin(A+B)=sinAcosB+cosAsinB\)

The only possibilites of \(\displaystyle A\) and \(\displaystyle B\) are 45 and 60 degrees interchangably. Substitute these values into the equation and evaluate.

\(\displaystyle sin(45+60)=sin45cos60+cos45sin60\)

\(\displaystyle sin(105)=(\frac{\sqrt2}{2})(\frac{1}{2})+(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})=\frac{\sqrt2+\sqrt6}{4}\)

Since \(\displaystyle \sin \alpha = \frac {1}{5}\) and \(\displaystyle \alpha\) is in quadrant I, we can say that \(\displaystyle y = 1\) and \(\displaystyle r = 5\) and therefore: 

\(\displaystyle x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}\). 

So \(\displaystyle \cos \alpha = \frac {2\sqrt{6}}{5}\).

Since \(\displaystyle \cos \beta = \frac {1}{4}\) and \(\displaystyle \beta\) is in quadrant I, we can say that \(\displaystyle x = 1\) and \(\displaystyle r = 4\) and therefore: 

\(\displaystyle y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}\). 

So \(\displaystyle \sin \beta = \frac {\sqrt{15}}{4}\).

Using the sine sum formula, we see:

\(\displaystyle \\ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} + \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 + 6 \sqrt {10}}{20}\)

Since \(\displaystyle \sin \alpha = \frac {1}{5}\) and \(\displaystyle \alpha\) is in quadrant I, we can say that \(\displaystyle y = 1\) and \(\displaystyle r = 5\) and therefore: 

\(\displaystyle x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}\). 

So \(\displaystyle \cos \alpha = \frac {2\sqrt{6}}{5}\).

Since \(\displaystyle \cos \beta = \frac {1}{4}\) and \(\displaystyle \beta\) is in quadrant I, we can say that \(\displaystyle x = 1\) and \(\displaystyle r = 4\) and therefore: 

\(\displaystyle y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}\). 

So \(\displaystyle \sin \beta = \frac {\sqrt{15}}{4}\).

Using the sine difference formula, we see:

\(\displaystyle \\ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} - \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 - 6 \sqrt {10}}{20}\)

\(\displaystyle \frac{5 \pi }{12}\) is equivalent to \(\displaystyle \frac{2 \pi }{12 } + \frac{3 \pi }{12 }\) or more simplified \(\displaystyle \frac{\pi}{6 } + \frac{\pi }{4}\).

We can use the sum identity to evaluate this sine:

\(\displaystyle \sin \left(\frac{\pi }{6} + \frac{\pi }{4 } \right) = \sin\left (\frac{\pi }{6} \right) \cos \left(\frac { \pi }{4} \right) + \cos \left(\frac{\pi }{6}\right) \sin \left( \frac{ \pi }{4} \right)\)

From the unit circle, we can determine these measures:

\(\displaystyle \frac{1}{2} \cdot \frac{1}{\sqrt 2 } + \frac{ \sqrt 3 }{2} \cdot \frac{1}{ \sqrt 2 } = \frac{1}{2 \sqrt 2 }+ \frac{\sqrt 3 }{2 \sqrt 2 } = \frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }\)

The angle \(\displaystyle 285 ^o = 225^ o + 60 ^ o\) or \(\displaystyle 45^o + 240 ^ o\).

Using the first one: 

\(\displaystyle \sin(60^o + 225^o ) = \sin (60^o) \cos (225^o) + \cos (60^o) \cos(225^o )\)

We can find these values in the unit circle:

\(\displaystyle \frac{\sqrt 3 }{2 } \cdot \frac{-1}{\sqrt 2 } + \frac{1}{2} \cdot \frac{-1}{\sqrt 2 } = \frac{-\sqrt3 }{2 \sqrt 2 } + \frac{-1}{2\sqrt2} = \frac{-\sqrt 3 - 1 }{2 \sqrt 2 }\)