I need to find the exact value of tan(arccos(-sqrt 11/6)).
To find the exact value of tan(arccos(-sqrt(11/6))), we can start by using the inverse trigonometric identity:
cos(arccos(x)) = x
In this case, we have arccos(-sqrt(11/6)). So, applying the identity, we get:
cos(arccos(-sqrt(11/6))) = -sqrt(11/6)
Now, recall the trigonometric identity:
tan^2(x) + 1 = sec^2(x)
Using this identity, we can rewrite the expression using the fact that sec(x) = 1/cos(x):
tan^2(arccos(-sqrt(11/6))) + 1 = sec^2(arccos(-sqrt(11/6)))
Plugging in the value we found earlier, we get:
tan^2(arccos(-sqrt(11/6))) + 1 = sec^2(arccos(-sqrt(11/6))) = (1 / cos(arccos(-sqrt(11/6))))^2 = (1 / -sqrt(11/6))^2
Now, simplifying further:
tan^2(arccos(-sqrt(11/6))) + 1 = sec^2(arccos(-sqrt(11/6))) = (1 / -sqrt(11/6))^2 = (1 / -sqrt(11/6))^2 = (6/11)
Taking the square root of both sides:
tan(arccos(-sqrt(11/6))) = sqrt(6/11)
Therefore, the exact value of tan(arccos(-sqrt(11/6))) is sqrt(6/11).
To find the exact value of tan(arccos(-√(11/6))), we can use the relationship between trigonometric functions and inverse trigonometric functions.
First, let's find the angle whose cosine is -√(11/6). We can express this angle using the arccosine function, denoted as arccos. So, we have:
θ = arccos(-√(11/6)).
To find the value of θ, we need to evaluate arccos(-√(11/6)). Since the cosine function returns values between 0 and π, or 0 and 180°, we can use the principal value of the inverse cosine function.
Next, to find the value of tan(θ), we use the tangent function, denoted as tan. Recall that tan(θ) = sin(θ) / cos(θ).
To calculate the tangent of an angle, we need the values of sine and cosine for that angle. However, finding the exact values of sine and cosine in this case could be complex since we are dealing with a non-standard angle.
Hence, we need to rely on a trigonometric identity to simplify the expression. One such identity is the Pythagorean Identity, which states:
sin²(θ) + cos²(θ) = 1.
We can rewrite this as:
sin²(θ) = 1 - cos²(θ).
Since we know the value of cos(θ) from the initial calculation of arccos(-√(11/6)), we can substitute it into the equation and solve for sin(θ). Once we have both sine and cosine values, we can finally calculate the tangent of the angle.
I hope this explanation helps you understand the process of finding the exact value of tan(arccos(-√(11/6))).