The expansion of Cos(A-B) is:
[tex]\text{Cos(A}-B)=CosACosB+SinASinB[/tex]
We are provided with the following:
[tex]\text{Cos A=}\frac{8}{17},Sin\text{ B=}\frac{3}{5}[/tex]
We will have to obtain the values of Cos B and Sin A. Thus, we have:
To be obtain Sin A, we have to get the value of the third side, which is the opposite side, by applying the pythagoras theorem. Thus, we have:
[tex]\begin{gathered} (\text{Hypotenuse)}^2=(\text{Opposite)}^2+(\text{Adjacent)}^2 \\ 17^2=O^2+8^2 \\ 289=O^2+64 \\ 289-64=O^2 \\ O^2=225 \\ O=\sqrt[]{225} \\ O=15 \\ \text{Thus, Sin A=}\frac{Opposite}{\text{Hypotenuse}} \\ Sin\text{ A=}\frac{15}{17} \end{gathered}[/tex]
To be obtain Cos B, we have to get the value of the third side, which is the adjacent side, by applying the pythagoras theorem. Thus, we have:
[tex]\begin{gathered} \text{Hyp}^2=\text{Opp}^2+\text{Adj}^2 \\ 5^2=3^2+A^2 \\ 25=9+A^2 \\ 25-9=A^2 \\ A^2=16 \\ A=\sqrt[]{16} \\ A=4 \\ \text{Thus Cos B=}\frac{Adjacent\text{ }}{\text{Hypotensue}} \\ \text{Cos B=}\frac{4}{5} \end{gathered}[/tex]
Now that we have obtained the values of Cos B and Sin A, we can then go on to solve the original problem.
[tex]\begin{gathered} \text{Cos(A}-B)=\text{CosACosB}+\text{SinASinB} \\ Cos(A-B)=\mleft\lbrace\frac{8}{17}\times\frac{4}{5}\mright\rbrace+\mleft\lbrace\frac{15}{17}\times\frac{3}{5}\mright\rbrace \\ \text{Cos(A-B)=}\frac{32}{85}+\frac{45}{85}_{} \\ \text{Cos(A}-B)=\frac{77}{85} \end{gathered}[/tex]
