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Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (90°, 3\sqrt{3}3), (135°,1), (180°,13\frac{1}{\sqrt{3}}31), (270°, 0)}.
y=cot(x3)y=\cot\left(\frac{x}{3}\right)y=cot(3x)
y=cot(x2)y=\cot\left(\frac{x}{2}\right)y=cot(2x)
y=cot(3x)y=\cot\left(3x\right)y=cot(3x)
Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (30°,−3-\sqrt{3}−3), (45°, -1), (60°, −13\frac{-1}{\sqrt{3}}3−1), (90°, 0)}.
y=cot(−x)y=\cot\left(-x\right)y=cot(−x)
y=cot(2x)y=\cot\left(2x\right)y=cot(2x)
Identify cotangent function from given set of ordered pairs: {(30°, ∞\infty∞), (60°,3\sqrt{3}3), (75°,1), (90°,13\frac{1}{\sqrt{3}}31), (120°, 0)}.
y=cot(x)y=\cot\left(x\right)y=cot(x)
y=cot(x−30°)y=\cot\left(x-30°\right)y=cot(x−30°)
Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (30°, 434\sqrt{3}43), (45°,4), (60°,43\frac{4}{\sqrt{3}}34), (90°, 0)}.
y=4cot(x)y=4\cot\left(x\right)y=4cot(x)
Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (60°, 3\sqrt{3}3), (90°,1), (120°,13\frac{1}{\sqrt{3}}31), (180°, 0)}.
Identify cotangent function from given set of ordered pairs: {(-30°, ∞\infty∞), (0°,−3-\sqrt{3}−3), (15°, -1), (30°,−13\frac{-1}{\sqrt{3}}3−1), (60°, 0)}.
y=−cot(x+30°)y=-\cot\left(x+30°\right)y=−cot(x+30°)
Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (15°,−3-\sqrt{3}−3), (22.5°, -1), (30°, −13\frac{-1}{\sqrt{3}}3−1), (45°, 0)}.
y=cot(−2x)y=\cot\left(-2x\right)y=cot(−2x)
y=cot(x+30°)y=\cot\left(x+30°\right)y=cot(x+30°)
Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (15°, 232\sqrt{3}23), (22.5°,2), (30°,23\frac{2}{\sqrt{3}}32), (45°, 0)}.
y=2cot(2x)y=2\cot\left(2x\right)y=2cot(2x)
Identify cotangent function from given set of ordered pairs: {(-30°, ∞\infty∞), (0°, 3\sqrt{3}3), (15°,1), (30°,13\frac{1}{\sqrt{3}}31), (60°, 0)}.
Identify cotangent function from given set of ordered pairs: {(0°, ∞\infty∞), (15°,3\sqrt{3}3), (22.5°,1), (30°,13\frac{1}{\sqrt{3}}31), (45°, 0)}.
It is done.