Question

find sinθ. cotθ = - 2/5, θ in quadrant iv sinθ = □ (simplify your answer, including any radicals. use integers or fractions

Explanation:

Step1: Use the Pythagorean identity \(1+\cot^{2}\theta=\csc^{2}\theta\)

Given \(\cot\theta =-\frac{2}{5}\), then \(\cot^{2}\theta=\frac{4}{25}\).
Substitute into the identity: \(1 +\frac{4}{25}=\csc^{2}\theta\).
So, \(\csc^{2}\theta=\frac{25 + 4}{25}=\frac{29}{25}\).

Step2: Find \(\csc\theta\)

Since \(\csc^{2}\theta=\frac{29}{25}\), then \(\csc\theta=\pm\frac{\sqrt{29}}{5}\).
Because \(\theta\) is in quadrant IV and \(\csc\theta=\frac{1}{\sin\theta}\), and \(\sin\theta<0\) in quadrant IV, so \(\csc\theta=-\frac{\sqrt{29}}{5}\).

Step3: Find \(\sin\theta\)

Since \(\sin\theta=\frac{1}{\csc\theta}\), then \(\sin\theta=-\frac{5}{\sqrt{29}}\).
Rationalize the denominator: \(\sin\theta=-\frac{5\sqrt{29}}{29}\).

Answer:

\(-\frac{5\sqrt{29}}{29}\)