How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$? [duplicate]

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  $cos(arcsin(x)) = sqrt1 - x^2$. How?
  
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How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?

I remember having to draw a triangle, but I'm not sure anymore.

trigonometry

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edited Aug 4 at 2:02

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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
  – Mike Pierce
  Aug 4 at 1:33
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

This question already has an answer here:

• 
  $cos(arcsin(x)) = sqrt1 - x^2$. How?
  
  7 answers
  
  

How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?

I remember having to draw a triangle, but I'm not sure anymore.

trigonometry

share|cite|improve this question

edited Aug 4 at 2:02

Blue

43.6k868141

asked Aug 4 at 1:30

Bas bas

18911

marked as duplicate by Simply Beautiful Art, Lord Shark the Unknown, lab bhattacharjee trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.

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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
  – Mike Pierce
  Aug 4 at 1:33
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

This question already has an answer here:

• 
  $cos(arcsin(x)) = sqrt1 - x^2$. How?
  
  7 answers
  
  

How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?

I remember having to draw a triangle, but I'm not sure anymore.

trigonometry

share|cite|improve this question

edited Aug 4 at 2:02

Blue

43.6k868141

asked Aug 4 at 1:30

Bas bas

18911

This question already has an answer here:

• 
  $cos(arcsin(x)) = sqrt1 - x^2$. How?
  
  7 answers
  
  

How to show that $cos(sin^-1(x))$ is $sqrt1-x^2$?

I remember having to draw a triangle, but I'm not sure anymore.

This question already has an answer here:

• 
  $cos(arcsin(x)) = sqrt1 - x^2$. How?
  
  7 answers
  
  

trigonometry

share|cite|improve this question

edited Aug 4 at 2:02

Blue

43.6k868141

asked Aug 4 at 1:30

Bas bas

18911

share|cite|improve this question

share|cite|improve this question

edited Aug 4 at 2:02

Blue

43.6k868141

edited Aug 4 at 2:02

Blue

43.6k868141

edited Aug 4 at 2:02

Blue

43.6k868141

43.6k868141

asked Aug 4 at 1:30

Bas bas

18911

asked Aug 4 at 1:30

Bas bas

18911

asked Aug 4 at 1:30

Bas bas

18911

18911

marked as duplicate by Simply Beautiful Art, Lord Shark the Unknown, lab bhattacharjee trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.

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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Simply Beautiful Art, Lord Shark the Unknown, lab bhattacharjee trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.

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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
  – Mike Pierce
  Aug 4 at 1:33
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
  – Mike Pierce
  Aug 4 at 1:33
  
  

  
  
  
  

2

2

The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
– Mike Pierce
Aug 4 at 1:33

The function $sin()$ takes a ratio, $x/1$ in this case, and returns an angle $theta = sin(x/1)$. Draw a triangle that models this, and then say what $cos(theta)$ is.
– Mike Pierce
Aug 4 at 1:33

add a comment | 

2 Answers
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accepted

Given $cos(sin^-1x)$

Let $sin^-1x=theta$
$$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$

Therefore, $cos(sin^-1x)=sqrt1-x^2$

share|cite|improve this answer

answered Aug 4 at 1:48

Key Flex

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add a comment | 

up vote
4
down vote

$cos(sin^-1(x)) = cos(sin^-1(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$

share|cite|improve this answer

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

answered Aug 4 at 1:44

Bas bas

18911

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Given $cos(sin^-1x)$

Let $sin^-1x=theta$
$$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$

Therefore, $cos(sin^-1x)=sqrt1-x^2$

share|cite|improve this answer

answered Aug 4 at 1:48

Key Flex

3,663422

add a comment | 

up vote
2
down vote



accepted

Given $cos(sin^-1x)$

Let $sin^-1x=theta$
$$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$

Therefore, $cos(sin^-1x)=sqrt1-x^2$

share|cite|improve this answer

answered Aug 4 at 1:48

Key Flex

3,663422

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Given $cos(sin^-1x)$

Let $sin^-1x=theta$
$$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$

Therefore, $cos(sin^-1x)=sqrt1-x^2$

share|cite|improve this answer

answered Aug 4 at 1:48

Key Flex

3,663422

Given $cos(sin^-1x)$

Let $sin^-1x=theta$
$$impliessintheta=x\sin^2theta=x^2\1-cos^2theta=x^2\cos^2theta=1-x^2\costheta=pmsqrt1-x^2\theta=cos^-1pmsqrt1-x^2$$
Now, plug in $theta=cos^-1pmsqrt1-x^2$ in $cos(sin^-1x)$ and we get$$pmsqrt1-x^2$$But note that $sin^-1x$ is in $left[dfrac-pi2,dfracpi2right]$ so $cos(sin^-1x)ge0$

Therefore, $cos(sin^-1x)=sqrt1-x^2$

share|cite|improve this answer

answered Aug 4 at 1:48

Key Flex

3,663422

share|cite|improve this answer

share|cite|improve this answer

answered Aug 4 at 1:48

Key Flex

3,663422

answered Aug 4 at 1:48

Key Flex

3,663422

answered Aug 4 at 1:48

Key Flex

3,663422

3,663422

add a comment | 

add a comment | 

up vote
4
down vote

$cos(sin^-1(x)) = cos(sin^-1(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$

share|cite|improve this answer

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

answered Aug 4 at 1:44

Bas bas

18911

add a comment | 

up vote
4
down vote

$cos(sin^-1(x)) = cos(sin^-1(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$

share|cite|improve this answer

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

answered Aug 4 at 1:44

Bas bas

18911

add a comment | 

up vote
4
down vote

up vote
4
down vote

$cos(sin^-1(x)) = cos(sin^-1(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$

share|cite|improve this answer

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

answered Aug 4 at 1:44

Bas bas

18911

$cos(sin^-1(x)) = cos(sin^-1(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $sqrt1^2 - x^2$

share|cite|improve this answer

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

answered Aug 4 at 1:44

Bas bas

18911

share|cite|improve this answer

share|cite|improve this answer

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

edited Aug 4 at 2:22

Simply Beautiful Art

49k571169

49k571169

answered Aug 4 at 1:44

Bas bas

18911

answered Aug 4 at 1:44

Bas bas

18911

answered Aug 4 at 1:44

Bas bas

18911

18911

add a comment | 

add a comment |