Question

Rewrite the expression cos(tan-1(x)) as an algebraic expression in x

Answer 1

To rewrite cos(tan^-1(x)) as an algebraic expression in x, we use a right-triangle approach to get the expression 1 / √(1 + x^2).

Step-by-step explanation:

To rewrite the expression cos(tan-1(x)) as an algebraic expression in x, we will use a right-triangle representation of trigonometric functions. Imagine a right triangle where the angle opposite 'x' is the tan-1(x), which means that the opposite side is x and the adjacent side is 1 (since tan is the ratio of the opposite side to the adjacent side).

By the Pythagorean theorem, the hypotenuse (h) can be represented as h = √(1 + x2). Now cos(tan-1(x)) is the adjacent side over the hypotenuse, leading to the algebraic expression cos(tan-1(x)) = 1 / √(1 + x2).

Answer 2

`$\cos\left(\tan^(-1) x\right)$` can be expressed algebraically as
`$√(1 - x^2)$`.

Let's use the trigonometric identity
`$\tan^(-1) x = (\pi)/(2) - \cos^(-1) x$` to rewrite the given expression
`$\cos\left(\tan^(-1) x\right)$` in terms of
`$x$`:

`\cos\left(\tan^(-1) x\right) &= \cos\left((\pi)/(2) - \cos^(-1) x\right)`

Now, we can use the cosine of the difference identity, which states that
`$\cos(a - b) = \cos a \cos b + \sin a \sin b$`.

In this case, let
`$a = (\pi)/(2)$` and
`$b = \cos^(-1) x$`:

`\cos\left((\pi)/(2) - \cos^(-1) x\right) &= \cos\left((\pi)/(2)\right) \cos\left(\cos^(-1) x\right) + \sin\left((\pi)/(2)\right) \sin\left(\cos^(-1) x\right)\\&= 0 \cdot \cos^(-1) x + 1 \cdot \sin\left(\cos^(-1) x\right)\\&= \sin\left(\cos^(-1) x\right)`

Finally, using the fact that
`$\sin(\cos^(-1) x) = √(1 - x^2)$`, the expression becomes:

`\sin\left(\cos^(-1) x\right) &= √(1 - x^2)`