Prove that cos(sin inverse x) = square root ( 1-x^2)
To prove that cos(sin^(-1)(x)) is equal to √(1 - x^2), we will use the concept of trigonometric identities.
We begin by considering a right-angled triangle. Assume that one of the angles within the triangle has a sine value of x. By definition, sin^(-1)(x) (or arcsin(x)) represents the angle whose sine is x.
Let's call this angle θ. We can write:
sin(θ) = x.
Now, let's label the sides of the triangle:
Opposite side = x (using the sine function definition)
Adjacent side = 1 (since this side is always the hypotenuse of the triangle)
Using the Pythagorean theorem, we can find the length of the hypotenuse (denoted as h):
h^2 = (Opposite side)^2 + (Adjacent side)^2
= x^2 + 1
Taking the square root of both sides:
h = √(x^2 + 1)
Since cos(θ) is defined as the ratio of the adjacent side to the hypotenuse, we have:
cos(θ) = Adjacent side / Hypotenuse
= 1 / √(x^2 + 1)
Simplifying the expression by multiplying both the numerator and the denominator by √(x^2 + 1), we get:
cos(θ) = (1 / √(x^2 + 1)) * (√(x^2 + 1) / √(x^2 + 1))
= √(x^2 + 1) / √(x^2 + 1)
= √(x^2 + 1) / (x^2 + 1)
Using the equation h = √(x^2 + 1), we can rewrite the expression as:
cos(θ) = √(x^2 + 1) / h
Since h is the hypotenuse, we can substitute h with 1:
cos(θ) = √(x^2 + 1) / 1
= √(x^2 + 1)
Therefore, we have proved that cos(sin^(-1)(x)) is equal to √(1 - x^2).