Answer:
Given that the Pythagorean identity states that
[tex]\begin{gathered} sin^2x+cos^2x=1 \\ and,\sin\theta=\frac{7}{10},find,\cos\theta \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} s\imaginaryI n^2x+cos^2x=1 \\ (\frac{7}{10})^2+cos^2\theta=1 \\ \frac{49}{100}+cos^2\theta=1 \end{gathered}[/tex]Collect similar terms by subtracting 49/100 from both sides
[tex]\begin{gathered} \frac{49}{100}+cos^{2}\theta=1 \\ \frac{49}{100}-\frac{49}{100}+cos^2\theta=1-\frac{49}{100} \\ cos^2\theta=\frac{100-49}{100} \\ cos^2\theta=\frac{51}{100} \end{gathered}[/tex]Square root both sides to calculate the value of cos θ
[tex]\begin{gathered} cos^{2}\theta=\frac{51}{100} \\ \cos\theta=\sqrt{\frac{51}{100}} \\ \cos\theta=\frac{\sqrt{51}}{10} \end{gathered}[/tex]Hence,
The value of cos θ is
[tex]\Rightarrow\cos\theta=\frac{\sqrt{51}}{10}[/tex]
