Question

9. Given the point (6,-8) values of the six trig function.

Answer 1

Answer:

Here's what I get.

Step-by-step explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10  

$\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}$

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

$\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}$