Question

QUICK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Find the exact values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (-6,6) lies on its terminal side.

Answer 1

x = -6, y = 6, Now find the hypotenuse:

(-6)² + (6)² = hypotenuse²

36 + 36 = hypotenuse²

   2(36) = hypotenuse²

 √2(36) = hypotenuse

      6√2 = hypotenuse

$sin\ \theta=\dfrac{y}{hypotenuse}=\dfrac{6}{6\sqrt2}=\dfrac{1}{\sqrt2}\cdot \dfrac{\sqrt2}{\sqrt2}=\boxed{\dfrac{\sqrt2}{2}}\\\\\\cos\ \theta=\dfrac{y}{hypotenuse}=\dfrac{-6}{6\sqrt2}=-\dfrac{1}{\sqrt2}\cdot \dfrac{\sqrt2}{\sqrt2}=\boxed{-\dfrac{\sqrt2}{2}}\\\\\\tan\ \theta=\dfrac{y}{x}=\dfrac{-6}{6}=\boxed{-1}\\\\\\csc\ \theta=\dfrac{hypotenuse}{y}=\dfrac{6\sqrt2}{6}=\boxed{\sqrt2}\\\\\\sec\ \theta=\dfrac{hypotenuse}{x}=\dfrac{6\sqrt2}{-6}=\boxed{-\sqrt2}$

$cot\ \theta=\dfrac{x}{y}=\dfrac{6}{-6}=\boxed{-1}$