How do you write an algebraic expression that is equivalent to $cos (arcsin (\frac{x - h}{r}))$ $cos(arcsin((x-h)/r))$ ?

Question

How do you write an algebraic expression that is equivalent to $cos (arcsin (\frac{x - h}{r}))$ $cos(arcsin((x-h)/r))$ ?

2 Answers

Impact of this question

8261 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-30T16:45:11

$cos (arcsin (\frac{x - h}{r})) = \sqrt{1 - {(\frac{x - h}{r})}^{2}}$ $cos(arcsin((x-h)/r)) = sqrt (1-((x-h)/r)^2)$

Explanation:

By definition:

$y = arcsin (\frac{x - h}{r})$ $y=arcsin((x-h)/r)$

is a number such that:

$sin y = \frac{x - h}{r}$ $siny = (x-h)/r$

Then we have that:

$\sqrt{1 - {sin}^{2} y} = cos y$ $sqrt(1-sin^2y) = cosy$

and thus:

$cos (arcsin (\frac{x - h}{r})) = \sqrt{1 - {(\frac{x - h}{r})}^{2}}$ $cos(arcsin((x-h)/r)) = sqrt (1-((x-h)/r)^2)$

Answer 2 · 2016-12-30T16:56:04

Use the identity $cos (θ) = \pm \sqrt{1 - {sin}^{2} (θ)}$ $cos(theta) = +-sqrt(1 - sin^2(theta))$ and then simplify.

Explanation:

We substitute ${sin}^{- 1} (\frac{x - h}{r})$ $sin^-1((x-h)/r)$ for $θ$ $theta$ :

$cos ({sin}^{- 1} (\frac{x - h}{r})) = \pm \sqrt{1 - {sin}^{2} ({sin}^{- 1} (\frac{x - h}{r}))}$ $cos(sin^-1((x-h)/r)) = +-sqrt(1 - sin^2(sin^-1((x - h)/r)))$

A property of a function and its inverse, $f (f^{- 1} (x)) = x$ $f(f^-1(x)) = x$ , makes the sine and the inverse sine disappear:

$cos ({sin}^{- 1} (\frac{x - h}{r})) = \pm \sqrt{1 - {(\frac{x - h}{r})}^{2}}$ $cos(sin^-1((x-h)/r)) = +-sqrt(1 - ((x - h)/r)^2)$

Make a common denominator:

$cos ({sin}^{- 1} (\frac{x - h}{r})) = \pm \sqrt{\frac{r^{2}}{r^{2}} - \frac{{(x - h)}^{2}}{r^{2}}}$ $cos(sin^-1((x-h)/r)) = +-sqrt(r^2/r^2 - (x - h)^2/r^2)$

$cos ({sin}^{- 1} (\frac{x - h}{r})) = \pm \sqrt{\frac{r^{2} - {(x - h)}^{2}}{r^{2}}}$ $cos(sin^-1((x-h)/r)) = +-sqrt((r^2 - (x - h)^2)/r^2)$

$cos ({sin}^{- 1} (\frac{x - h}{r})) = \pm \frac{\sqrt{r^{2} - {(x - h)}^{2}}}{r}$ $cos(sin^-1((x-h)/r)) = +-sqrt(r^2 - (x - h)^2)/r$

Science

Math

Humanities

... and beyond

How do you write an algebraic expression that is equivalent to $cos (arcsin (\frac{x - h}{r}))$ $cos(arcsin((x-h)/r))$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you write an algebraic expression that is equivalent to cos(arcsin(x−hr))cos(arcsin((x-h)/r))?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you write an algebraic expression that is equivalent to $cos (arcsin (\frac{x - h}{r}))$ $cos(arcsin((x-h)/r))$ ?