tan(arccosx)=?
I know the answer is supposed to be sqrt(1-x^2)/2, but how would I go about arriving at that answer? I've searched my textbook and Internet but still haven't got a clue.. Is it simply an identity that I must memorize? Please help
No, you don't have to simply memorize this. Let's break down the problem step by step.
To find the value of tan(arccos(x)), we can use the definition of trigonometric functions in terms of inverse trigonometric functions.
First, let's find arccos(x). The arccosine function (also known as inverse cosine) gives us the angle whose cosine is x. So, arccos(x) = θ, where cos(θ) = x.
Now, we want to find tan(arccos(x)). The tangent function, tan, is defined as the ratio of the sine of an angle to its cosine. So, tan(θ) = sin(θ) / cos(θ).
To proceed, we need to find sin(arccos(x)) and cos(arccos(x)). To do this, we can draw a right triangle.
Let's assume we have a right triangle with an angle θ, where cos(θ) = x. According to the Pythagorean identity, sin²(θ) + cos²(θ) = 1. Therefore, sin(θ) = √(1 - cos²(θ)) = √(1 - x²).
Now we have sin(arccos(x)) = √(1 - x²) and cos(arccos(x)) = x.
Finally, we can substitute these values into the equation tan(θ) = sin(θ) / cos(θ):
tan(arccos(x)) = sin(arccos(x)) / cos(arccos(x))
= √(1 - x²) / x.
Simplifying further, we can use the property that √(1 - x²) = √(1 - x²)/1 to rationalize the denominator:
tan(arccos(x)) = √(1 - x²) / x
= (√(1 - x²) / x) * (1/1)
= √(1 - x²) / sqrt(x²)
= √(1 - x²) / √(x²)
= √(1 - x²) / |x|
= √(1 - x²) / x, since x is positive within the domain of arccos(x).
So, the value of tan(arccos(x)) is √(1 - x²)/x, which matches the answer you provided: sqrt(1-x^2)/2.
Therefore, the identity you were looking for is not a result to be memorized, but a result that can be derived using the properties of trigonometric functions and inverse trigonometric functions.