angle theta is in the standard position and 0'=<theta=<180'. If cos theta = -2/3, find the exact value of sin theta* tan theta.
You know that cosθ = x/r
In QII, x < 0 and y > 0, so we know
x = -2
r = 3
so, y = √5
Now you know that
sinθ = √5/3
and you now can easily figure sinθ * tanθ
oh ok thanks a lot for your help :D
To find the exact value of sin theta * tan theta when cos theta is known, we can use the trigonometric identity: sin^2 theta + cos^2 theta = 1.
From this identity, we can solve for sin theta:
sin^2 theta = 1 - cos^2 theta
sin theta = ± sqrt(1 - cos^2 theta)
Given that cos theta = -2/3, we can substitute this value into the equation:
sin theta = ± sqrt(1 - (-2/3)^2)
sin theta = ± sqrt(1 - 4/9)
sin theta = ± sqrt(5/9)
Now, to find tan theta, we can use the identity: tan theta = sin theta / cos theta. Substituting the values we found earlier:
tan theta = (± sqrt(5/9)) / (-2/3)
tan theta = ± (sqrt(5/9) * (-3/2))
tan theta = ± (-3/2) * (sqrt(5/9))
tan theta = ± (-3 * sqrt(5)) / (2 * sqrt(9))
tan theta = ± (-3 * sqrt(5)) / (2 * 3)
tan theta = ± (-sqrt(5) / 2)
Finally, to find the value of sin theta * tan theta, we simply multiply the two values:
sin theta * tan theta = (± sqrt(5/9)) * (± (-sqrt(5) / 2))
sin theta * tan theta = (± sqrt(5) / sqrt(9)) * (± (-sqrt(5) / 2))
sin theta * tan theta = (± sqrt(5) / 3) * (± (-sqrt(5) / 2))
sin theta * tan theta = (± -5/3) / 2
sin theta * tan theta = ± (-5/6)
Therefore, the exact value of sin theta * tan theta is ± (-5/6).