Applying the next trigonometric identity:
[tex]\sin ^2\alpha+\cos ^2\alpha=1[/tex]to angle θ, and solving for cos(θ), we get:
[tex]\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ \text{ Substituting with }\sin ^{}\theta=1/3\colon \\ (\frac{1}{3})^2+\cos ^2\theta=1 \\ \frac{1}{9}+\cos ^2\theta=1 \\ \cos ^2\theta=1-\frac{1}{9} \\ \cos ^2\theta=\frac{8}{9} \\ \cos \theta=\sqrt[]{\frac{8}{9}} \\ \cos \theta=\frac{\sqrt[]{8}}{3} \\ \cos \theta=\frac{2\sqrt[]{2}}{3} \end{gathered}[/tex]The relation between the cosine and the secant of an angle is:
[tex]\begin{gathered} \sec \varphi=\frac{1}{\cos\varphi} \\ \text{ Substituting with }\sec \varphi=17/8\colon \\ \frac{17}{8}=\frac{1}{\cos\varphi} \\ (\frac{17}{8})^{-1}=(\frac{1}{\cos\varphi})^{-1} \\ \frac{8}{17}=\cos \varphi \end{gathered}[/tex]Applying the before mentioned trigonometric identity to angle φ, we get:
[tex]\begin{gathered} \sin ^2\varphi+\cos ^2\varphi=1 \\ \text{ Substituting with cos}\varphi\text{ = 8/17}\colon \\ \sin ^2\varphi+(\frac{8}{17})^2=1 \\ \sin ^2\varphi+\frac{64}{289}^{}=1 \\ \sin ^2\varphi=1-\frac{64}{289} \\ \sin ^2\varphi=\frac{225}{289} \\ \sin \varphi=\sqrt[]{\frac{225}{289}} \\ \sin \varphi=\frac{15}{17} \end{gathered}[/tex]Difference formula of cosine
[tex]\cos (\alpha-\beta)=\cos \alpha\cdot\cos \beta+\sin \alpha\cdot\sin \beta[/tex]Applying this formula to angles θ and φ, and substituting with the values found, we get:
[tex]\begin{gathered} \cos (\theta-\varphi)=\cos \theta\cdot\cos \varphi+\sin \theta\cdot\sin \varphi \\ \cos (\theta-\varphi)=\frac{2\sqrt[]{2}}{3}\cdot\frac{8}{17}+\frac{1}{3}\cdot\frac{15}{17} \\ \cos (\theta-\varphi)=\frac{16\sqrt[]{2}}{51}+\frac{15}{51} \\ \cos (\theta-\varphi)=\frac{16\sqrt[]{2}+15}{51} \end{gathered}[/tex]