Given:
[tex]cos\text{ }s=\frac{1}{5};sin\text{ }t=\frac{4}{5}[/tex]
We will find cos (s+t) and cos (s-t)
First, we need to find the sin (s) and cos (t) using the trigonometric Pythagorean identity: sin²x + cos²x = 1
So,
[tex]\begin{gathered} sin\text{ }s=\sqrt{1-cos^2s}=\frac{\sqrt{24}}{5} \\ \\ cos\text{ }t=\sqrt{1-sin^2t}=\frac{3}{5} \end{gathered}[/tex]
Second, we will find cos(s+t) using the difference identity as follows:
[tex]cos(s+t)=cos\text{ }s*cos\text{ }t-sin\text{ }s*sin\text{ }t[/tex]
Substitute with the values of the sine and cosine of both angles.
[tex]cos(s+t)=\frac{1}{5}*\frac{3}{5}-\frac{\sqrt{24}}{5}*\frac{4}{5}=\frac{3}{25}-\frac{4\sqrt{24}}{25}=\frac{3-8\sqrt{6}}{25}[/tex]
Finally, we will find cos (s - t)
[tex]cos(s-t)=cos\text{ }s*cos\text{ }t+sin\text{ }s*sin\text{ }t[/tex]
Substitute with the values of the sine and cosine of both angles.
[tex]cos(s-t)=\frac{1}{5}*\frac{3}{5}+\frac{\sqrt{24}}{5}*\frac{4}{5}=\frac{3}{25}+\frac{8\sqrt{6}}{25}=\frac{3+8\sqrt{6}}{25}[/tex]
So, the answer will be:
[tex]\begin{gathered} cos(s+t)=\frac{3-8\sqrt{6}}{25} \\ \\ cos(s-t)=\frac{3+8\sqrt{6}}{25} \end{gathered}[/tex]
