Question

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

Answer 1

cotΘ = -
`√(2)`

Explanation:

Using the trigonometric identities

cot x =
`(cosx)/(sinx)`

sin²x + cos²x = 1 ⇒ cosx = ±
`√(1-sin^2x)`

Since Θ is in second quadrant then cosΘ < 0

cosΘ = -
`\sqrt{1-((√(3) )/(3) }`)^2

= -
`\sqrt{1-(1)/(3) }` = -
`\sqrt{(2)/(3) }` = -
`(√(2) )/(√(3) )`

Hence

cotΘ =
`(-(√(2) )/(√(3) ) )/((√(3) )/(3) )`

= -
`(√(2) )/(√(3) )` ×
`(3)/(√(3) )`

= -
`(3√(2) )/(3)` = -
`√(2)`