Question

What is cot θ when sin θ = square root of 3 divided by 3? Rationalize the denominator if necessary.

Answer 1

Answer:

cotΘ = - $\sqrt{2}$

Step-by-step explanation:

Using the trigonometric identities

cot x = $\frac{cosx}{sinx}$

sin²x + cos²x = 1 ⇒ cosx = ± $\sqrt{1-sin^2x}$

Since Θ is in second quadrant then cosΘ < 0

cosΘ = - $\sqrt{1-(\frac{\sqrt{3} }{3} }$)^2

        = - $\sqrt{1-\frac{1}{3} }$ = - $\sqrt{\frac{2}{3} }$ = - $\frac{\sqrt{2} }{\sqrt{3} }$

Hence

cotΘ = $\frac{-\frac{\sqrt{2} }{\sqrt{3} } }{\frac{\sqrt{3} }{3} }$

        = - $\frac{\sqrt{2} }{\sqrt{3} }$ × $\frac{3}{\sqrt{3} }$

        = - $\frac{3\sqrt{2} }{3}$ = - $\sqrt{2}$