find the exact value of each of the remaining five trigonometric functions of the acute angle theta if sine = square root 6/6

Question

sin theta = square root of 6/6

Answer 1

Solution:

Given: sin (θ) = √6 /6 , and sin(θ) = opposite / hypotenuse of a right angle triangle with θ the acute angle between the adjacent and the hypotenuse. Knowing that in a right angle triangle:

hypotenuse² = opposite² + adjacent², it follows that the adjacent² = hypotenuse² - opposite²

= (6²) - (√6)² = 36 - 6 = 30 and adjacent = √30

cos(θ) = adjacent / hypotenuse = √30 / 6

tan(θ) = sin(θ) / cos(θ) = (√6 /6) / (√30 / 6) = (√6 /6) ( 6/√30) = √6 / √30

cot(θ) = 1 / tan(θ) = cos(θ) / sin (θ) = (√30 / 6) / (√6 /6) = (√30 / 6) / (6 / √6) = √30 / √6

sec(θ) = 1 /cos(θ) = 1 / (√30 / 6) = 6 / √30

csc(θ) = 1 / sin (θ) = 1 / (√6 /6) = 6 / √6