Use a right triangle to write the following expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x

Question

Use a right triangle to write the following expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. cot(sin^-1x)

Answer 1

First, draw a right triangle ABC where C is the right angle. I drew angle A smaller than angle B, no particular reason. I'll call the opposite side of angle A "a", opposite side of angle B "b", and so on. So c² = a² + b².

Let x = a / c. Then sin^-1x = A (the angle), and cot(sin^-1x) = cotA = b / a.

Now using the Pythagorean theorem, b = √(c² - a²).

Substitute this into cotA = b / a, and it becomes cotA = (√(c²-a²)) / a = √((c/a)²-1).

Well, x was a / c. This means our (c/a)² is just 1/x².

Substituting 1/x² in the place of (c/a)², and we finally get the algebraic expression of cot(sin^-1x) as √(1/x² - 1). Alternatively, we can write that as (√(1 - x²))/x.

Answer 2

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|a \c

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|________X_\_

b

SinX = a/c

Sin^-1(a/c) = X

CotX = b/a