use a right triangle to write the expression as an algebraic expression. assume that x is positive and in the domain of the given inverse trigonometric function.
sin(sec^-1 �ãx^2+ 25/(x))
the symbol before x^2 should be a square root
To begin, let's visualize a right triangle to help us write the expression as an algebraic expression. In a right triangle, we have three sides: the opposite side (o), the adjacent side (a), and the hypotenuse (h).
Now, let's consider the inverse trigonometric function given in the question, sec^(-1) (x^2 + 25/x). In a right triangle with angle A, the secant function (sec) is equal to the hypotenuse over the adjacent side, or h/a. So we can write:
sec(A) = h/a
In this case, the angle A is sec^(-1) (x^2 + 25/x).
Next, we'll use the property of trigonometric functions that states: sin^2(A) + cos^2(A) = 1. Rearranging this equation, we have:
sin^2(A) = 1 - cos^2(A)
Substituting sec(A) = 1/cos(A) into this equation, we get:
sin^2(A) = 1 - (1/cos^2(A))
Since sin(A) = o/h and cos(A) = a/h, we can rewrite the equation as:
(o/h)^2 = 1 - (1/(a/h))^2
Simplifying further, we have:
o^2/h^2 = 1 - (h^2/a^2)
Now, let's substitute the values given in the question: o = x^2 + 25/x, a = x, and h = √(x^2 + (x^2 + 25/x)^2).
Substituting these values into the equation, we get:
(x^2 + 25/x)^2 / (√(x^2 + (x^2 + 25/x)^2))^2 = 1 - (√(x^2 + (x^2 + 25/x)^2)/x)^2
Simplifying the expression, we have:
(x^2 + 25/x)^2 / (x^2 + (x^2 + 25/x)^2) = 1 - (x^2 + (x^2 + 25/x)^2)/x^2
Therefore, the algebraic expression equivalent to sin(sec^(-1) (x^2 + 25/x)) is:
(x^2 + 25/x)^2 / (x^2 + (x^2 + 25/x)^2) = 1 - (x^2 + (x^2 + 25/x)^2)/x^2