Write the following trigonometric expression as an algebraic expression in x free of trigonometric or inverse trigonometric functions
sin(cos^-1 x) -1≤x≤1
draw a triangle with adjacent/hypotenuse = x/1
the other leg is √(1-x^2)
so, sin(arccos(x)) = √(1-x^2)
or, if θ = arccos(x), then we have the ubiquitous identity
sin^2θ + cos^2θ = 1
but cosθ = x, so
sin^2θ + x^2 = 1
sin^2θ = 1-x^2
sinθ = √(1-x^2)
To rewrite the trigonometric expression as an algebraic expression in x, we can make use of the relationship between cosine and sine functions.
The expression sin(cos^-1 x) represents the sine of the inverse cosine of x. Since the range of the inverse cosine function is limited to [0, π], and the sine function operates within that domain, we can use the following identity:
sin(cos^-1 x) = √(1 - x^2)
This identity is derived from the Pythagorean identity sin^2θ + cos^2θ = 1, where θ is the angle resulting from the inverse cosine operation.
Therefore, the algebraic expression in x, free of trigonometric or inverse trigonometric functions, is:
√(1 - x^2) for -1 ≤ x ≤ 1