find the cosine for an angle that has a sine of 3√40/20 and is in quadrant II. use the pythagorean identity sin^2 (0) + cos^2 (0) = 1 and the quadrant to solve
If the sine of an angle θ is \(\frac{3\sqrt{40}}{20}\), then by simplifying the fraction, you get:
\[ \sin(θ) = \frac{3\sqrt{40}}{20} = \frac{3\sqrt{4 \cdot 10}}{20} = \frac{3 \cdot 2\sqrt{10}}{20} = \frac{6\sqrt{10}}{20} = \frac{3\sqrt{10}}{10}. \]
Now using the Pythagorean identity:
\[ \sin^2(θ) + \cos^2(θ) = 1 \]
Plug in \(\sin(θ) = \frac{3\sqrt{10}}{10}\) into the formula to find \(\cos^2(θ)\):
\[ \left(\frac{3\sqrt{10}}{10}\right)^2 + \cos^2(θ) = 1 \]
\[ \frac{9 \cdot 10}{100} + \cos^2(θ) = 1 \]
\[ \frac{90}{100} + \cos^2(θ) = 1 \]
\[ \cos^2(θ) = 1 - \frac{90}{100} \]
\[ \cos^2(θ) = \frac{100}{100} - \frac{90}{100} \]
\[ \cos^2(θ) = \frac{10}{100} \]
\[ \cos^2(θ) = \frac{1}{10} \]
To find \(\cos(θ)\), we take the square root of both sides, keeping in mind that since θ is in the second quadrant, the cosine should be negative (because cosine is negative in the second quadrant):
\[ \cos(θ) = -\sqrt{\frac{1}{10}} \]
Therefore, the cosine for an angle that has a sine of \(\frac{3\sqrt{40}}{20}\) and is in the second quadrant would be:
\[ \cos(θ) = -\frac{1}{\sqrt{10}} = -\frac{\sqrt{10}}{10}. \]