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=(y)/(hypotenuse)=(6)/(6\sqrt2)=(1)/(\sqrt2)\cdot (\sqrt2)/(\sqrt2)=\boxed{(\sqrt2)/(2)}\\\\\\cos\ \theta=(y)/(hypotenuse)=(-6)/(6\sqrt2)=-(1)/(\sqrt2)\cdot (\sqrt2)/(\sqrt2)=\boxed{-(\sqrt2)/(2)}\\\\\\tan\ \theta=(y)/(x)=(-6)/(6)=\boxed{-1}\\\\\\csc\ \theta=(hypotenuse)/(y)=(6\sqrt2)/(6)=\boxed{\sqrt2}\\\\\\sec\ \theta=(hypotenuse)/(x)=(6\sqrt2)/(-6)=\boxed{-\sqrt2}`

`cot\ \theta=(x)/(y)=(6)/(-6)=\boxed{-1}`