cos(sin^-1 x)=sq rt(1-x^2)
prove the identity
draw a triangle with opposite leg=x, hypotenuse=1
adjacent leg = √(1-x^2)
however, doing it algebraically,
let θ=sin^-1 x
x = sinθ
cos^2 θ + sin^2 θ = 1
cos^2 θ + x^2 = 1
cos^2 θ = 1 - x^2
cosθ = √(1-x^2)
To prove the identity cos(sin^-1 x) = √(1 - x^2), we will use the geometric interpretation of trigonometric functions.
Let's consider a right triangle where one of the acute angles is denoted as θ. The side opposite to θ is represented by x, and the hypotenuse is represented by 1 (by assuming the hypotenuse as 1, we can then use ratios to express the sides relative to the hypotenuse).
We can start by expressing sinθ using the triangle sides:
sinθ = opposite / hypotenuse = x / 1 = x
Now, using the Pythagorean theorem, we can calculate the adjacent side:
adjacent^2 = hypotenuse^2 - opposite^2
adjacent^2 = 1^2 - x^2
adjacent^2 = 1 - x^2
Now, we can find cosθ by dividing the adjacent side by the hypotenuse:
cosθ = adjacent / hypotenuse = √(1 - x^2) / 1 = √(1 - x^2)
Therefore, cos(sin^-1 x) = √(1 - x^2), which proves the given identity.