Cos theta= 1/3, find the remaining trigonometric functions of the acute angle theta
To find the remaining trigonometric functions of the acute angle theta, we need to first determine the values of the other functions using the given value of cos(theta).
Given that cos(theta) = 1/3, we can use the Pythagorean identity: sin^2(theta) + cos^2(theta) = 1.
Plugging in the value of cos(theta) = 1/3, we have:
sin^2(theta) + (1/3)^2 = 1
sin^2(theta) + 1/9 = 1
sin^2(theta) = 1 - 1/9
sin^2(theta) = 8/9
Taking the square root of both sides, we get:
sin(theta) = sqrt(8/9)
sin(theta) = sqrt(8)/sqrt(9)
sin(theta) = 2sqrt(2)/3
Now, let's find the tangent of theta. We know that:
tan(theta) = sin(theta)/cos(theta)
Plugging in the values we found, we have:
tan(theta) = (2sqrt(2)/3) / (1/3)
tan(theta) = (2sqrt(2)/3) * (3/1)
tan(theta) = 2sqrt(2)
So, the remaining trigonometric functions of the acute angle theta are:
sin(theta) = 2sqrt(2)/3
cos(theta) = 1/3
tan(theta) = 2sqrt(2)
draw the triangle. the missing side (opposite Theta) is sqrt(9-1)=sqrt8
Is there any reason you cannot write out the missing functions?