Question

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

Answer 1

Answer:

sin θ = $\frac{\sqrt{6} }{3}$

Step-by-step explanation:

Given that,

cos θ = $\frac{\sqrt{3} }{3}$

From the trigonometric functions,

cos θ = $\frac{adjacent}{hypotenus}$

⇒ adjacent = $\sqrt{3}$, and the hypotenuse = 3

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

$/hyp/^{2}$ = $/adj/^{2}$ + $/opp/^{2}$

$/3/^{2}$ = $(\sqrt{3} )^{2}$ + $x^{2}$

9 = 3 + $x^{2}$

$x^{2}$ = 9 - 3

   = 6

x = $\sqrt{6}$

Thus, opposite side = $\sqrt{6}$

So that,

sin θ = $\frac{opposite}{hypotenuse}$

        = $\frac{\sqrt{6} }{3}$

Therefore,

sin θ = $\frac{\sqrt{6} }{3}$