given that $\\sin\\theta = \\frac{45}{53}$ and $\\cos\\theta>0$, determine the values of the sine and cosine…

Explanation:

Step1: Find (\cos\theta)

Use the identity (\sin^{2}\theta+\cos^{2}\theta = 1). Given (\sin\theta=\frac{45}{53}), then (\cos^{2}\theta=1-\sin^{2}\theta). [ \begin{align*} \cos^{2}\theta&=1 - (\frac{45}{53})^{2}\ &=\frac{53^{2}-45^{2}}{53^{2}}\ &=\frac{(53 - 45)(53 + 45)}{53^{2}}\ &=\frac{8\times98}{53^{2}}\ &=\frac{784}{2809}\ \cos\theta&=\frac{28}{53}\quad(\text{since }\cos\theta>0) \end{align*} ]

Step2: Find (\cos2\theta)

Use the double - angle formula (\cos2\theta=1 - 2\sin^{2}\theta) (or (\cos2\theta=2\cos^{2}\theta-1)). Using (\cos2\theta=1 - 2\sin^{2}\theta): [ \begin{align*} \cos2\theta&=1-2\times(\frac{45}{53})^{2}\ &=1-\frac{2\times2025}{2809}\ &=\frac{2809-4050}{2809}\ &=\frac{2809 - 4050}{2809}\ &=\frac{28^{2}-45^{2}}{53^{2}}\ &=\frac{784 - 2025}{2809}\ &=\frac{- 1241}{2809} \end{align*} ] Using (\cos2\theta=2\cos^{2}\theta-1): [ \begin{align*} \cos2\theta&=2\times(\frac{28}{53})^{2}-1\ &=\frac{2\times784}{2809}-1\ &=\frac{1568}{2809}-\frac{2809}{2809}\ &=\frac{1568 - 2809}{2809}\ &=\frac{- 1241}{2809} \end{align*} ]

Answer:

(\cos2\theta=-\frac{1241}{2809})