Consider this equation. cos(theta) = (-2√5)/5. If theta is an angle in quadrant II, what is the value of sin(theta)? NO LINKS!!!
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[tex]\\ \tt\hookrightarrow sin^2\theta=1-cos^2\theta[/tex]
[tex]\\ \tt\hookrightarrow sin^2\theta=1-(\dfrac{-2\sqrt{5}}{5})^2[/tex]
[tex]\\ \tt\hookrightarrow sin^2\theta=1-\dfrac{20}{25}[/tex]
[tex]\\ \tt\hookrightarrow sin^2\theta=\dfrac{5}{25}[/tex]
[tex]\\ \tt\hookrightarrow sin\theta=\dfrac{\sqrt{5}}{5}[/tex]
[tex]\\ \tt\hookrightarrow sin\theta=\dfrac{1}{\sqrt{5}}[/tex]