tan(3arc cos3/4)?
=3cos3/4
=cos=4
sqrt17/4 = tan
arccos3/4, the angle whose cosine is 3/4 is 41.4 degrees. Three times that angle is 124.2 degrees.
The tangent of that angle is -1.467
I don't know what your last three equations are about. They don't make mathematical sense.
To find the value of tan(3arccos(3/4)), we can start by using the identity:
tan(2θ) = (2tanθ)/(1 - tan^2θ)
First, we need to find the value of arccos(3/4). Let's call it θ.
arccos(3/4) = θ
To find the value of θ, we'll use the inverse cosine function and evaluate it with 3/4:
θ = arccos(3/4)
Next, we substitute this value of θ into the identity for tan(2θ):
tan(2θ) = (2tanθ)/(1 - tan^2θ)
tan(2θ) = (2tan(arccos(3/4)))/(1 - tan^2(arccos(3/4)))
Now, we can use the Pythagorean identity for tanθ to find the value of tan(arccos(3/4)):
tan(arccos(3/4)) = √(1 - (cos(arccos(3/4)))^2) / cos(arccos(3/4))
Since cos(arccos(3/4)) = 3/4, we can simplify this expression:
tan(arccos(3/4)) = √(1 - (3/4)^2) / (3/4)
tan(arccos(3/4)) = √(1 - 9/16) / (3/4)
tan(arccos(3/4)) = √(7/16) / (3/4)
Simplifying further:
tan(arccos(3/4)) = √7 / 3
Now, we substitute this value back into the identity for tan(2θ):
tan(2θ) = (2 √7 / 3)/(1 - (√7 / 3)^2)
Simplifying further:
tan(2θ) = (2 √7 / 3)/(1 - 7/9)
tan(2θ) = (2 √7 / 3)/(2/9)
tan(2θ) = (2 √7 / 3) * (9/2)
Canceling out common factors:
tan(2θ) = 3 √7
Therefore, tan(3arccos(3/4)) = 3 √7.