Question

Cos(0)=square root 3/3, sin 0<0. what is the value of sin0​

Answer 1

sin θ =
`(√(6) )/(3)`

Explanation:

Given that,

cos θ =
`(√(3) )/(3)`

From the trigonometric functions,

cos θ =
`(adjacent)/(hypotenus)`

⇒ adjacent =
`√(3)`, and the hypotenuse = 3

Let the opposite side be represented by x, applying the Pythagoras theorem we have;

`/hyp/^(2)` =
`/adj/^(2)` +
`/opp/^(2)`

`/3/^(2)` =
`(√(3) )^(2)` +
`x^(2)`

9 = 3 +
`x^(2)`

`x^(2)` = 9 - 3

= 6

x =
`√(6)`

Thus, opposite side =
`√(6)`

So that,

sin θ =
`(opposite)/(hypotenuse)`

=
`(√(6) )/(3)`

Therefore,

sin θ =
`(√(6) )/(3)`