Simple square root factoring question

  • Thread starter Odyssey
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  • #1
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Hi,

is [tex]\sqrt{a^2-a^2\sin^2{x}} = a\cos{x}?[/tex]

If not, what should it be? :confused:

Appreciate the help! :smile:
 
Last edited:

Answers and Replies

  • #2
574
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That is correct.
 
  • #3
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Thank you for the help Sirus!
 
  • #4
Tide
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It is correct within a sign!
 
  • #5
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Tide said:
It is correct within a sign!
What do you mean?
 
  • #6
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Sirus said:
That is correct.
No, that is not quite correct if you think of how a secondary definition of the modulus (absolute value) follows from the square root of a square,

[tex]\sqrt{x^2} = \|x\|[/tex]

In your case

[tex]\sqrt{a^2-a^2\sin^2{x}} = a\cos{x}[/tex]

can be written if and only if a and cos(x) are both positive or both negative; before you brought them out of the square root sign, you had the intermediate step,

[tex]\sqrt{a^2-a^2\sin^2{x}} = \sqrt{a^2\cos^2{x}}[/tex]

so

[tex]\sqrt{a^2-a^2\sin^2{x}} = \sqrt{a^2\cos^2{x}} = \|a\cos x\|[/tex]

which should hold true anyway since the left hand side is the positive square root.

In general however, you can write it as a cos x if you have no major problems with the signs (you won't have any if theta lies between 0 and pi/2 and a > 0 for instance). But if you're proving something which involves this substitution, you had rather take it into account.

Cheers
Vivek
 
  • #7
574
2
I stand corrected. I forgot about that.
 

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